Timer: Off
Random: Off
401. What does “mode” in a waveguide refer to?
ⓐ. The frequency of the signal only
ⓑ. The path of rays in the medium
ⓒ. The field distribution pattern of electromagnetic waves inside the guide
ⓓ. The amplitude variation of the wave
Correct Answer: The field distribution pattern of electromagnetic waves inside the guide
Explanation: Modes describe the spatial distribution of electric and magnetic fields that satisfy boundary conditions in a waveguide. Different modes correspond to different cutoff frequencies and propagation characteristics.
402. In a rectangular waveguide, which is the dominant mode?
ⓐ. TE$_{01}$
ⓑ. TE$_{10}$
ⓒ. TM$_{01}$
ⓓ. TEM
Correct Answer: TE$_{10}$
Explanation: The TE$_{10}$ mode has the lowest cutoff frequency and therefore propagates most efficiently. It is the dominant mode for rectangular waveguides, meaning it is the first mode to appear as frequency increases.
403. Why can’t TEM mode exist in a hollow rectangular waveguide?
ⓐ. Because TEM requires both electric and magnetic fields to be zero
ⓑ. Because TEM requires two conductors for field lines to terminate
ⓒ. Because waveguides cannot support any modes
ⓓ. Because TEM modes only exist in free space
Correct Answer: Because TEM requires two conductors for field lines to terminate
Explanation: TEM modes need both electric and magnetic fields perpendicular to propagation with terminating boundaries. Hollow waveguides with a single conducting wall cannot provide this; hence only TE and TM modes are supported.
404. What is the cutoff frequency formula for TE$_{mn}$ mode in rectangular waveguides?
ⓐ. $f_c = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$
ⓑ. $f_c = \frac{c}{\lambda}$
ⓒ. $f_c = \frac{c}{2a}$
ⓓ. $f_c = \frac{c}{ab}$
Correct Answer: $f_c = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$
Explanation: For a rectangular waveguide with dimensions $a$ and $b$, cutoff frequency depends on mode indices $m,n$. TE and TM modes have cutoff frequencies based on these indices, determining whether a mode propagates.
405. For a circular waveguide, what does the cutoff frequency of a mode depend on?
ⓐ. Radius and Bessel function root values
ⓑ. Length of waveguide only
ⓒ. Dielectric constant only
ⓓ. Aperture angle
Correct Answer: Radius and Bessel function root values
Explanation: Cutoff frequencies in circular waveguides are derived using Bessel functions. The mode indices and the corresponding root values of Bessel functions, along with the radius of the waveguide, define the cutoff.
406. What happens if a wave is operated below its cutoff frequency in a waveguide?
ⓐ. It propagates normally
ⓑ. It gets absorbed completely
ⓒ. It becomes evanescent and decays exponentially
ⓓ. It generates TEM waves
Correct Answer: It becomes evanescent and decays exponentially
Explanation: Below cutoff, modes cannot satisfy the propagation condition. Instead of transmitting energy, the fields decay exponentially along the length of the waveguide, making transmission impossible.
407. In rectangular waveguides, the cutoff frequency of TE$_{20}$ mode is how many times that of TE$_{10}$ mode?
ⓐ. Equal
ⓑ. Twice
ⓒ. Half
ⓓ. Four times
Correct Answer: Twice
Explanation: For TE$_{10}$: $f_c = \frac{c}{2a}$.
For TE$_{20}$: $f_c = \frac{c}{a}$.
Thus, cutoff frequency of TE$_{20}$ is exactly twice TE$_{10}$.
408. Which mode is dominant in circular waveguides?
ⓐ. TE$_{01}$
ⓑ. TE$_{11}$
ⓒ. TM$_{11}$
ⓓ. TEM
Correct Answer: TE$_{11}$
Explanation: The TE$_{11}$ mode has the lowest cutoff frequency in circular waveguides. Hence, it propagates first and is called the dominant mode, similar to TE$_{10}$ in rectangular guides.
409. Why is mode analysis important in waveguides?
ⓐ. To increase length of waveguides
ⓑ. To determine which frequency ranges can propagate efficiently
ⓒ. To avoid resonance
ⓓ. To reduce reflection completely
Correct Answer: To determine which frequency ranges can propagate efficiently
Explanation: Each mode has a cutoff frequency below which it does not propagate. Mode analysis helps engineers choose appropriate waveguide dimensions and frequency ranges for efficient operation.
410. A rectangular waveguide has $a = 3 \, \text{cm}, b = 1 \, \text{cm}$. Calculate cutoff frequency of TE$_{10}$ mode (take $c = 3 \times 10^8 \, \text{m/s}$).
ⓐ. 3 GHz
ⓑ. 5 GHz
ⓒ. 6 GHz
ⓓ. 10 GHz
Correct Answer: 5 GHz
Explanation: Formula: $f_c = \frac{c}{2a}$.
Here, $a = 0.03 \, \text{m}$.
$f_c = \frac{3 \times 10^8}{2 \times 0.03}$.
$f_c = \frac{3 \times 10^8}{0.06}$.
$f_c = 5.0 \times 10^9 \, \text{Hz} = 5 \, \text{GHz}$.
411. Why are waveguides preferred over coaxial cables in microwave engineering?
ⓐ. Waveguides are cheaper than cables
ⓑ. Waveguides have lower losses and higher power-handling capability at high frequencies
ⓒ. Waveguides support TEM mode only
ⓓ. Waveguides are easier to bend than cables
Correct Answer: Waveguides have lower losses and higher power-handling capability at high frequencies
Explanation: At microwave frequencies, coaxial cables suffer from significant conductor and dielectric losses. Waveguides, due to their hollow structure and mode confinement, exhibit lower attenuation and can handle higher microwave power, making them ideal for radar and satellite applications.
412. Which mode is commonly used in rectangular waveguides for microwave transmission?
ⓐ. TE$_{01}$
ⓑ. TE$_{10}$
ⓒ. TM$_{01}$
ⓓ. TEM
Correct Answer: TE$_{10}$
Explanation: The TE$_{10}$ mode is the dominant mode in rectangular waveguides because it has the lowest cutoff frequency and therefore propagates most efficiently with minimal loss in microwave systems.
413. Which of the following devices uses waveguides in microwave engineering?
ⓐ. Magnetron
ⓑ. Radar antenna feeds
ⓒ. Microwave ovens
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Magnetrons produce microwaves which are guided through waveguides, radar antennas use waveguides for efficient power transfer, and microwave ovens use waveguides to direct microwave energy into the cooking cavity.
414. In optical fibers, why is light confined within the core?
ⓐ. Because of diffraction only
ⓑ. Because of total internal reflection at the core–cladding boundary
ⓒ. Because fiber has metallic walls
ⓓ. Because light speed is zero in cladding
Correct Answer: Because of total internal reflection at the core–cladding boundary
Explanation: The refractive index of the core is slightly higher than that of the cladding. Light entering at an angle below the critical angle is totally internally reflected, confining it to the fiber and enabling long-distance communication.
415. Which type of fiber is used for long-distance, high-speed communication?
ⓐ. Multimode step-index fiber
ⓑ. Multimode graded-index fiber
ⓒ. Single-mode fiber
ⓓ. Plastic fiber
Correct Answer: Single-mode fiber
Explanation: Single-mode fibers allow only one propagation mode, eliminating modal dispersion. This ensures high bandwidth and minimal signal loss, making them ideal for long-distance, high-speed communication systems.
416. What is the main limitation caused by dispersion in optical fibers?
ⓐ. Increase in refractive index
ⓑ. Broadening of pulses leading to inter-symbol interference
ⓒ. Increase in signal amplitude
ⓓ. Decrease in frequency of operation
Correct Answer: Broadening of pulses leading to inter-symbol interference
Explanation: Dispersion spreads the optical pulse over time, causing overlapping between adjacent pulses in digital communication. This overlap (inter-symbol interference) reduces the ability to distinguish transmitted signals accurately.
417. A light pulse at wavelength $1550 \, \text{nm}$ travels through fiber with dispersion coefficient $D = 16 \, \text{ps}/(\text{nm·km})$. For $\Delta \lambda = 2 \, \text{nm}$ over 50 km, calculate pulse broadening.
ⓐ. 1.2 ns
ⓑ. 1.6 ns
ⓒ. 2.0 ns
ⓓ. 2.5 ns
Correct Answer: 1.6 ns
Explanation: Formula: $\Delta T = D \cdot \Delta\lambda \cdot L$.
Here, $D = 16, \Delta\lambda = 2, L = 50$.
$\Delta T = 16 \times 2 \times 50$.
$\Delta T = 1600 \, \text{ps} = 1.6 \, \text{ns}$.
418. Which microwave component often uses rectangular waveguides?
ⓐ. Inductors
ⓑ. Resonant cavities
ⓒ. Diodes
ⓓ. Transformers
Correct Answer: Resonant cavities
Explanation: Rectangular waveguides are used to build resonant cavities that store microwave energy at specific frequencies. These are used in oscillators, filters, and klystrons to stabilize and amplify microwave signals.
419. What is the advantage of graded-index fibers in communication systems?
ⓐ. They eliminate chromatic dispersion
ⓑ. They reduce modal dispersion by gradually varying refractive index
ⓒ. They are cheaper than single-mode fibers
ⓓ. They allow infinite bandwidth
Correct Answer: They reduce modal dispersion by gradually varying refractive index
Explanation: Graded-index fibers have a core whose refractive index decreases gradually from center to edge. This compensates for different modal paths, reducing dispersion and improving bandwidth compared to step-index multimode fibers.
420. Why are waveguides not used for low-frequency signals?
ⓐ. Because they cannot support TE modes
ⓑ. Because required dimensions become very large compared to wavelength
ⓒ. Because losses are extremely high at low frequencies
ⓓ. Because waveguides are non-conductive
Correct Answer: Because required dimensions become very large compared to wavelength
Explanation: The cutoff frequency of waveguides depends on their physical dimensions. At low frequencies with large wavelengths, the waveguide cross-section must be very large to support propagation, making them impractical compared to cables.
421. What is wave polarization?
ⓐ. The change of wavelength in a medium
ⓑ. The orientation of the electric field vector in an electromagnetic wave
ⓒ. The bending of waves around obstacles
ⓓ. The increase of frequency in guided media
Correct Answer: The orientation of the electric field vector in an electromagnetic wave
Explanation: Polarization describes the orientation of the electric field vector of an electromagnetic wave as it propagates. Depending on how the electric field varies with time, the wave can be linearly, circularly, or elliptically polarized.
422. In linear polarization, how does the electric field vector behave?
ⓐ. It rotates continuously with constant magnitude
ⓑ. It remains fixed in one direction while oscillating
ⓒ. It spirals along the direction of propagation
ⓓ. It changes amplitude randomly
Correct Answer: It remains fixed in one direction while oscillating
Explanation: In linear polarization, the electric field oscillates in a single fixed direction perpendicular to the propagation. This can be vertical or horizontal, and is commonly used in radio and optical communication.
423. Which of the following is true for circular polarization?
ⓐ. The electric field oscillates in one fixed plane
ⓑ. The electric field vector rotates with constant magnitude in a circular path
ⓒ. The electric field magnitude changes randomly
ⓓ. The magnetic field becomes zero
Correct Answer: The electric field vector rotates with constant magnitude in a circular path
Explanation: Circular polarization occurs when two perpendicular linear components of equal amplitude have a phase difference of $90^\circ$. The result is a rotating electric field vector tracing a circle in time.
424. Elliptical polarization is:
ⓐ. A special case of linear polarization
ⓑ. A general case where the electric field vector traces an ellipse
ⓒ. Not physically possible in electromagnetic waves
ⓓ. The same as circular polarization
Correct Answer: A general case where the electric field vector traces an ellipse
Explanation: Elliptical polarization occurs when the two perpendicular field components have unequal amplitudes and/or phase differences other than $90^\circ$. Linear and circular polarization are special cases of elliptical polarization.
425. Which type of polarization is used in most LCD displays?
ⓐ. Linear polarization
ⓑ. Circular polarization
ⓒ. Elliptical polarization
ⓓ. Random polarization
Correct Answer: Linear polarization
Explanation: LCDs use linear polarizers to control light intensity. The alignment of liquid crystal molecules changes the orientation of light polarization, controlling the passage of light through a second polarizer.
426. A wave has two perpendicular components of equal amplitude, but one lags behind the other by $90^\circ$. What type of polarization results?
ⓐ. Linear
ⓑ. Circular
ⓒ. Elliptical
ⓓ. Random
Correct Answer: Circular
Explanation: Circular polarization occurs when perpendicular electric field components are equal in magnitude and differ in phase by $\pm 90^\circ$. The tip of the electric field vector describes a circle.
427. A wave has two perpendicular components of unequal amplitude with a phase difference of $90^\circ$. What type of polarization is this?
ⓐ. Linear
ⓑ. Circular
ⓒ. Elliptical
ⓓ. Random
Correct Answer: Elliptical
Explanation: With unequal amplitudes, the rotating electric field vector does not form a circle but an ellipse. Elliptical polarization is the most general form, encompassing both linear and circular as special cases.
428. Which type of polarization is often used in satellite communication to reduce interference between channels?
ⓐ. Random polarization
ⓑ. Orthogonal linear polarization
ⓒ. Circular polarization
ⓓ. Elliptical polarization only
Correct Answer: Circular polarization
Explanation: Circular polarization ensures that signal orientation is maintained regardless of receiver alignment. It also allows use of left-hand and right-hand polarizations to separate signals and reduce interference.
429. If a wave is linearly polarized at $45^\circ$, what is the relation between its horizontal and vertical components?
ⓐ. Only horizontal exists
ⓑ. Only vertical exists
ⓒ. Horizontal and vertical components are equal in amplitude and in phase
ⓓ. Horizontal and vertical components are equal but $90^\circ$ out of phase
Correct Answer: Horizontal and vertical components are equal in amplitude and in phase
Explanation: A linearly polarized wave at $45^\circ$ has equal horizontal and vertical electric field components. Since they are in phase, the resultant vector oscillates along a line at $45^\circ$.
430. Why is polarization analysis important in optics and communication?
ⓐ. It increases the frequency of light
ⓑ. It helps in controlling and filtering electromagnetic waves
ⓒ. It reduces wavelength of light
ⓓ. It eliminates dispersion completely
Correct Answer: It helps in controlling and filtering electromagnetic waves
Explanation: Polarization analysis allows engineers to design filters, antennas, and optical devices that selectively transmit or block certain polarizations. This improves signal quality, reduces interference, and is widely used in communication, microscopy, and LCD technology.
431. An EM wave has horizontal and vertical components: $E_x = 2 \cos(\omega t)$, $E_y = 2 \cos(\omega t)$. What is the polarization of the wave?
ⓐ. Circular
ⓑ. Linear at $45^\circ$
ⓒ. Elliptical
ⓓ. Random
Correct Answer: Linear at $45^\circ$
Explanation: Here, both components have equal amplitude (2) and are in phase.
This results in a straight-line polarization at an angle of $45^\circ$ to the x-axis.
432. For a wave: $E_x = 3 \cos(\omega t)$, $E_y = 4 \cos(\omega t)$. Find the type of polarization.
ⓐ. Circular
ⓑ. Elliptical
ⓒ. Linear along x-axis
ⓓ. Linear along y-axis
Correct Answer: Elliptical
Explanation: Amplitudes are unequal (3 and 4), but the phase difference is zero.
This produces a linear polarization along a tilted axis, but in terms of the general definition, it is an ellipse collapsed into a line (special elliptical case).
433. An EM wave has $E_x = 5 \cos(\omega t)$, $E_y = 5 \cos(\omega t + 90^\circ)$. What is the polarization?
ⓐ. Linear
ⓑ. Circular
ⓒ. Elliptical
ⓓ. Random
Correct Answer: Circular
Explanation: Equal amplitudes in orthogonal components (5 and 5).
Phase difference = $90^\circ$.
This gives circular polarization, as the tip of the electric field vector traces a circle.
434. A wave has $E_x = 10 \cos(\omega t)$ and $E_y = 5 \cos(\omega t + 90^\circ)$. Find the polarization.
ⓐ. Linear
ⓑ. Circular
ⓒ. Elliptical
ⓓ. Random
Correct Answer: Elliptical
Explanation: Amplitudes are unequal (10 and 5).
Phase difference = $90^\circ$.
This gives elliptical polarization, since the electric field vector traces an ellipse.
435. A linearly polarized wave at $60^\circ$ has an electric field amplitude of $10 \, \text{V/m}$. Find the horizontal and vertical components.
ⓐ. $E_x = 5$, $E_y = 8.7$
ⓑ. $E_x = 8.7$, $E_y = 5$
ⓒ. $E_x = 7.0$, $E_y = 7.0$
ⓓ. $E_x = 10$, $E_y = 0$
Correct Answer: $E_x = 5$, $E_y = 8.7$
Explanation: Formula: $E_x = E \cos \theta$, $E_y = E \sin \theta$.
$E_x = 10 \cos 60^\circ = 10 \times 0.5 = 5$.
$E_y = 10 \sin 60^\circ = 10 \times 0.866 = 8.66 \approx 8.7$.
436. A wave has $E_x = 6 \cos(\omega t)$ and $E_y = 8 \cos(\omega t)$. What is the polarization angle with x-axis?
ⓐ. $30^\circ$
ⓑ. $45^\circ$
ⓒ. $53^\circ$
ⓓ. $60^\circ$
Correct Answer: $53^\circ$
Explanation: Polarization angle = $\tan^{-1}(E_y/E_x)$.
$\tan^{-1}(8/6) = \tan^{-1}(1.333)$.
Angle ≈ $53^\circ$.
437. An EM wave has $E_x = 7 \cos(\omega t)$, $E_y = 7 \cos(\omega t + 180^\circ)$. What is the polarization?
ⓐ. Linear horizontal
ⓑ. Circular
ⓒ. Linear vertical
ⓓ. Linear at an angle
Correct Answer: Linear at an angle
Explanation: Amplitudes equal (7,7).
Phase difference = $180^\circ$.
The wave becomes linear, polarized along a line at an angle, but opposite directions cancel one axis periodically.
438. A circularly polarized wave has frequency $2 \, \text{GHz}$. What is its period of field rotation?
ⓐ. 0.25 ns
ⓑ. 0.5 ns
ⓒ. 1.0 ns
ⓓ. 2.0 ns
Correct Answer: 0.5 ns
Explanation: Period = $T = 1/f$.
$T = 1/(2 \times 10^9)$.
$T = 0.5 \times 10^{-9} \, \text{s} = 0.5 \, \text{ns}$.
439. A wave has components $E_x = 2 \cos(\omega t)$, $E_y = 2 \cos(\omega t + 45^\circ)$. What is the polarization?
ⓐ. Linear
ⓑ. Circular
ⓒ. Elliptical
ⓓ. Random
Correct Answer: Elliptical
Explanation: Equal amplitudes (2 and 2).
Phase difference ≠ $0^\circ$ or $90^\circ$ (here $45^\circ$).
Thus, the electric field traces an ellipse.
440. A wave has linear polarization at $30^\circ$ to the x-axis, with amplitude $20 \, \text{V/m}$. Find its components.
ⓐ. $E_x = 10$, $E_y = 17.3$
ⓑ. $E_x = 17.3$, $E_y = 10$
ⓒ. $E_x = 15$, $E_y = 15$
ⓓ. $E_x = 20$, $E_y = 0$
Correct Answer: $E_x = 17.3$, $E_y = 10$
Explanation: $E_x = E \cos \theta = 20 \cos 30^\circ = 20 \times 0.866 = 17.32$.
$E_y = E \sin \theta = 20 \sin 30^\circ = 20 \times 0.5 = 10$.
441. What does polarization of light mean?
ⓐ. Restricting light to travel in only one direction
ⓑ. Restricting the vibrations of the electric field to a single direction
ⓒ. Increasing the speed of light in a medium
ⓓ. Splitting light into multiple wavelengths
Correct Answer: Restricting the vibrations of the electric field to a single direction
Explanation: Light is a transverse wave with electric and magnetic fields oscillating perpendicular to propagation. Polarization refers to restricting these oscillations to a single direction, which is not naturally the case in ordinary unpolarized light.
442. Which of the following can polarize light?
ⓐ. Reflection
ⓑ. Refraction
ⓒ. Scattering
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Light can be polarized by reflection (Brewster’s angle effect), refraction through birefringent materials, and scattering in the atmosphere. Each method restricts oscillations of the electric field to specific directions.
443. At what angle of incidence is reflected light completely polarized?
ⓐ. Critical angle
ⓑ. Brewster’s angle
ⓒ. Right angle
ⓓ. Zero degrees
Correct Answer: Brewster’s angle
Explanation: At Brewster’s angle, reflected and refracted rays are perpendicular. The reflected wave has its electric field entirely in the plane perpendicular to the plane of incidence, producing complete polarization.
444. A beam of unpolarized light of intensity $I_0$ passes through a polarizer. What is the intensity of transmitted light?
ⓐ. $I_0$
ⓑ. $I_0/2$
ⓒ. $2I_0$
ⓓ. Zero
Correct Answer: $I_0/2$
Explanation: A polarizer transmits only the component of electric field aligned with its axis. For unpolarized light, average transmission is 50%, so transmitted intensity is $I_0/2$.
445. Malus’ law states that:
ⓐ. $I = I_0 \cos \theta$
ⓑ. $I = I_0 \cos^2 \theta$
ⓒ. $I = I_0 \sin \theta$
ⓓ. $I = I_0 \sin^2 \theta$
Correct Answer: $I = I_0 \cos^2 \theta$
Explanation: According to Malus’ law, the transmitted intensity through an analyzer is proportional to the square of the cosine of the angle between incident polarization direction and the analyzer axis.
446. A plane-polarized light of intensity $20 \, \text{W/m}^2$ passes through an analyzer at $60^\circ$ to the plane of polarization. Find transmitted intensity.
ⓐ. 5 W/m$^2$
ⓑ. 10 W/m$^2$
ⓒ. 15 W/m$^2$
ⓓ. 20 W/m$^2$
Correct Answer: 5 W/m$^2$
Explanation: Formula: $I = I_0 \cos^2 \theta$.
Here, $I_0 = 20, \theta = 60^\circ$.
$I = 20 \cos^2 60^\circ$.
$\cos 60^\circ = 0.5$.
$I = 20 \times (0.25)$.
$I = 5 \, \text{W/m}^2$.
447. Which natural phenomenon demonstrates polarization by scattering?
ⓐ. Rainbows
ⓑ. Blue color of the sky
ⓒ. Red sunset
ⓓ. Double refraction
Correct Answer: Blue color of the sky
Explanation: Sunlight is scattered by air molecules in the atmosphere. Scattered light is partially polarized, and polarization is maximum at 90° to the Sun’s direction. This is why polarized sunglasses reduce glare from the sky.
448. A calcite crystal shows double images when placed over text. This phenomenon is due to:
ⓐ. Reflection
ⓑ. Refraction only
ⓒ. Birefringence
ⓓ. Diffraction
Correct Answer: Birefringence
Explanation: Calcite is a birefringent crystal, meaning it has two different refractive indices for orthogonal polarizations. This splits light into ordinary and extraordinary rays, producing double images.
449. If unpolarized light is incident on a polarizer, followed by another polarizer rotated at $45^\circ$, what fraction of original intensity is transmitted?
ⓐ. 1/2
ⓑ. 1/4
ⓒ. 1/8
ⓓ. 1
Correct Answer: 1/4
Explanation: First polarizer reduces intensity to $I_0/2$. Second polarizer transmits according to Malus’ law: $I = (I_0/2) \cos^2 45^\circ = (I_0/2)(0.5) = I_0/4$.
450. Which optical device uses polarization to reduce glare from surfaces like water and glass?
ⓐ. Lenses
ⓑ. Polarizing sunglasses
ⓒ. Diffraction gratings
ⓓ. Prisms
Correct Answer: Polarizing sunglasses
Explanation: Polarizing sunglasses use polarizing filters that block horizontally polarized light reflected from surfaces like water and roads. This reduces glare and improves visibility, especially in outdoor environments.
451. Why are polarizers used in cameras and sunglasses?
ⓐ. To amplify light intensity
ⓑ. To reduce glare by blocking specific polarizations of reflected light
ⓒ. To increase the speed of light in glass
ⓓ. To change the frequency of incident light
Correct Answer: To reduce glare by blocking specific polarizations of reflected light
Explanation: Reflected light from water, glass, or roads is partially polarized horizontally. Polarizing filters allow only vertical polarization to pass, reducing glare and enhancing clarity in vision and photography.
452. Why is polarization important in LCD technology?
ⓐ. It increases brightness by doubling intensity
ⓑ. It controls light transmission through liquid crystal orientation
ⓒ. It reduces wavelength of light
ⓓ. It eliminates reflection losses
Correct Answer: It controls light transmission through liquid crystal orientation
Explanation: LCDs use crossed polarizers. Light passes through the first polarizer, and liquid crystals rotate polarization depending on applied voltage. The second polarizer allows or blocks light depending on polarization orientation, forming images.
453. Which type of polarization is often used in satellite communication?
ⓐ. Random polarization
ⓑ. Vertical linear polarization only
ⓒ. Circular polarization
ⓓ. Elliptical polarization only
Correct Answer: Circular polarization
Explanation: Circular polarization ensures signals are received regardless of antenna alignment. Satellites often use right-hand and left-hand circular polarizations to transmit multiple channels without interference.
454. In fiber-optic communication, why is polarization management important?
ⓐ. It reduces signal amplitude
ⓑ. It prevents pulse broadening caused by polarization mode dispersion
ⓒ. It eliminates the need for amplifiers
ⓓ. It increases wavelength stability
Correct Answer: It prevents pulse broadening caused by polarization mode dispersion
Explanation: In single-mode fibers, different polarization modes can travel at slightly different velocities. This leads to polarization mode dispersion, broadening pulses and limiting data rates. Controlling polarization minimizes this effect.
455. A beam of unpolarized light passes through two polarizers at $30^\circ$ to each other. Find the transmitted intensity relative to the incident intensity $I_0$.
ⓐ. $I_0/2$
ⓑ. $I_0/4$
ⓒ. $I_0/8$
ⓓ. $I_0/16$
Correct Answer: $I_0/4$
Explanation: First polarizer transmits $I_0/2$.
Second polarizer transmits according to Malus’ law: $I = (I_0/2) \cos^2 30^\circ$.
$\cos^2 30^\circ = (0.866)^2 = 0.75$.
So, $I = (I_0/2) \times 0.75 = 0.375 I_0$.
Closest option = $I_0/4$.
456. Why are polarization filters used in 3D movie glasses?
ⓐ. To block light of certain colors only
ⓑ. To separate left-eye and right-eye images using orthogonal polarizations
ⓒ. To reduce the speed of light entering each eye
ⓓ. To increase resolution of the projector
Correct Answer: To separate left-eye and right-eye images using orthogonal polarizations
Explanation: 3D movies project two images polarized differently (horizontal/vertical or circular left/right). Glasses with orthogonal polarizers ensure each eye receives only one image, creating stereoscopic depth perception.
457. Why is polarization used in optical isolators in communication systems?
ⓐ. To allow light transmission in one direction only
ⓑ. To amplify the signal
ⓒ. To reduce the wavelength of transmitted light
ⓓ. To change linear polarization into circular
Correct Answer: To allow light transmission in one direction only
Explanation: Optical isolators use polarizers and Faraday rotators. They transmit light in one direction while blocking back-reflected light. This prevents feedback into lasers in communication systems, ensuring stable operation.
458. In optical fiber communication, polarization-maintaining fibers are designed to:
ⓐ. Increase chromatic dispersion
ⓑ. Keep light polarization constant during propagation
ⓒ. Eliminate total internal reflection
ⓓ. Reduce frequency of light signals
Correct Answer: Keep light polarization constant during propagation
Explanation: Standard fibers cause polarization states to change randomly due to stress and imperfections. Polarization-maintaining fibers have built-in asymmetry that keeps the polarization stable, which is essential in sensors and coherent communication systems.
459. Which optical device uses birefringence to control polarization states?
ⓐ. Beam splitter
ⓑ. Half-wave plate
ⓒ. Concave lens
ⓓ. Optical grating
Correct Answer: Half-wave plate
Explanation: A half-wave plate introduces a $180^\circ$ phase difference between orthogonal components of light. This rotates the polarization direction without altering intensity, making it useful in laser optics and communication systems.
460. Why is polarization important in wireless communication using antennas?
ⓐ. Polarization determines alignment and efficient energy transfer between transmitter and receiver
ⓑ. Polarization changes frequency of signals
ⓒ. Polarization eliminates diffraction losses
ⓓ. Polarization reduces antenna length
Correct Answer: Polarization determines alignment and efficient energy transfer between transmitter and receiver
Explanation: Maximum signal strength is achieved when transmitting and receiving antennas have the same polarization. A mismatch (e.g., vertical vs. horizontal) causes polarization loss, reducing efficiency in wireless and satellite communication.
461. What is a wavefront?
ⓐ. A line or surface representing points on a wave that are in the same phase
ⓑ. A line showing direction of particle vibration
ⓒ. A line showing energy transfer only
ⓓ. A boundary separating two different media
Correct Answer: A line or surface representing points on a wave that are in the same phase
Explanation: A wavefront is defined as the locus of points vibrating in the same phase. For example, in spherical waves, the wavefronts are concentric spheres; in plane waves, they are parallel planes.
462. What is a ray in wave physics?
ⓐ. A narrow beam of particles
ⓑ. A line drawn perpendicular to the wavefront showing direction of wave propagation
ⓒ. A line tangent to the wavefront
ⓓ. A vibrating path of particles
Correct Answer: A line drawn perpendicular to the wavefront showing direction of wave propagation
Explanation: Rays are imaginary lines drawn normal to the wavefronts, representing the path along which energy propagates. They are used in geometrical optics for simple analysis of wave motion.
463. In a plane progressive wave, the wavefront is:
ⓐ. Circular
ⓑ. Linear or planar
ⓒ. Elliptical
ⓓ. Spherical
Correct Answer: Linear or planar
Explanation: A plane progressive wave has parallel wavefronts extending infinitely in two dimensions. This occurs when waves from a distant source reach a small region, making the wavefronts effectively plane.
464. In the case of a point source emitting waves, the wavefronts are:
ⓐ. Plane
ⓑ. Circular (in 2D) or spherical (in 3D)
ⓒ. Elliptical
ⓓ. Linear
Correct Answer: Circular (in 2D) or spherical (in 3D)
Explanation: Waves from a point source spread uniformly in all directions. In two dimensions they form circular wavefronts, and in three dimensions, they form concentric spherical wavefronts.
465. Which of the following is correct about the relationship between rays and wavefronts?
ⓐ. Rays are always parallel to wavefronts
ⓑ. Rays are always perpendicular to wavefronts
ⓒ. Rays are sometimes parallel, sometimes perpendicular
ⓓ. Rays are tangent to wavefronts
Correct Answer: Rays are always perpendicular to wavefronts
Explanation: By definition, rays represent the direction of propagation of energy, while wavefronts represent surfaces of constant phase. The direction of energy transfer is always normal to the wavefront.
466. What kind of wavefronts do parallel rays of sunlight produce on Earth’s surface?
ⓐ. Circular wavefronts
ⓑ. Spherical wavefronts
ⓒ. Plane wavefronts
ⓓ. Elliptical wavefronts
Correct Answer: Plane wavefronts
Explanation: Since the Sun is very far away, its spherical wavefronts appear almost flat by the time they reach Earth. Hence, sunlight produces plane wavefronts on Earth’s surface.
467. Which wavefront is produced when a convex lens is illuminated by a point source at its focus?
ⓐ. Plane wavefront
ⓑ. Spherical converging wavefront
ⓒ. Spherical diverging wavefront
ⓓ. Elliptical wavefront
Correct Answer: Plane wavefront
Explanation: A convex lens converts spherical diverging waves from a point source at its focus into parallel rays. Since rays are parallel, the corresponding wavefronts are plane.
468. A concave mirror produces which type of wavefront when a parallel beam is incident on it?
ⓐ. Plane wavefront
ⓑ. Spherical converging wavefront
ⓒ. Spherical diverging wavefront
ⓓ. Circular wavefront
Correct Answer: Spherical converging wavefront
Explanation: A concave mirror focuses incident parallel rays to its focus. Therefore, the reflected rays form concentric spherical converging wavefronts.
469. In water waves produced by dropping a stone into a pond, what kind of wavefronts are observed?
ⓐ. Plane wavefronts
ⓑ. Circular wavefronts
ⓒ. Elliptical wavefronts
ⓓ. Random wavefronts
Correct Answer: Circular wavefronts
Explanation: The disturbance created spreads out in two dimensions on the water surface, forming concentric circular wavefronts with the point of impact as the center.
470. Which principle explains the formation and propagation of wavefronts?
ⓐ. Fermat’s principle
ⓑ. Huygens’ principle
ⓒ. Malus’ law
ⓓ. Doppler effect
Correct Answer: Huygens’ principle
Explanation: According to Huygens’ principle, every point on a wavefront acts as a secondary source of spherical wavelets. The envelope of these secondary wavelets gives the new wavefront, explaining the propagation of wavefronts.
471. What is the relationship between rays and wavefronts?
ⓐ. Rays are parallel to wavefronts
ⓑ. Rays are perpendicular to wavefronts
ⓒ. Rays are tangential to wavefronts
ⓓ. Rays have no relation with wavefronts
Correct Answer: Rays are perpendicular to wavefronts
Explanation: Rays represent the direction of energy propagation, while wavefronts represent surfaces of constant phase. Since energy moves outward normally, rays are always perpendicular to the wavefronts.
472. In a plane wave, what is the relation between rays and wavefronts?
ⓐ. Rays are diverging, wavefronts are spherical
ⓑ. Rays are parallel, wavefronts are plane
ⓒ. Rays are circular, wavefronts are elliptical
ⓓ. Rays are random, wavefronts are irregular
Correct Answer: Rays are parallel, wavefronts are plane
Explanation: For a plane wave, the wavefronts are flat and extend infinitely. Rays are drawn perpendicular to these planes, making them parallel to each other.
473. In spherical waves, rays and wavefronts are related as:
ⓐ. Rays are radial, wavefronts are concentric spheres
ⓑ. Rays are parallel, wavefronts are concentric circles
ⓒ. Rays are tangent, wavefronts are ellipses
ⓓ. Rays and wavefronts are random
Correct Answer: Rays are radial, wavefronts are concentric spheres
Explanation: In spherical waves, wavefronts are concentric spheres centered at the source. Rays are drawn radially outward from the source and are perpendicular to the spherical surfaces.
474. If rays from a distant source are nearly parallel, the corresponding wavefront is:
ⓐ. Spherical
ⓑ. Circular
ⓒ. Plane
ⓓ. Elliptical
Correct Answer: Plane
Explanation: A very distant source produces wavefronts that appear flat in a small region. Since rays are nearly parallel, the wavefront is effectively plane.
475. What happens to wavefronts and rays when light passes through a convex lens?
ⓐ. Diverging spherical wavefronts become plane wavefronts, rays parallel
ⓑ. Plane wavefronts become diverging spherical wavefronts, rays radial
ⓒ. Diverging spherical wavefronts remain diverging
ⓓ. Rays and wavefronts are unchanged
Correct Answer: Diverging spherical wavefronts become plane wavefronts, rays parallel
Explanation: A convex lens transforms diverging spherical wavefronts from a point source into plane wavefronts. This corresponds to parallel rays emerging from the lens.
476. In reflection from a plane mirror, how are incident and reflected rays related to the wavefront?
ⓐ. They are parallel to the wavefront
ⓑ. They remain perpendicular to incident and reflected wavefronts
ⓒ. They form random angles with wavefronts
ⓓ. They coincide with wavefront surfaces
Correct Answer: They remain perpendicular to incident and reflected wavefronts
Explanation: Both incident and reflected rays are normal to their respective wavefronts. The law of reflection ensures the angles between rays and the mirror are consistent with perpendicularity to wavefronts.
477. Which principle mathematically relates rays and wavefronts in geometrical optics?
ⓐ. Huygens’ principle
ⓑ. Fermat’s principle
ⓒ. Doppler principle
ⓓ. Principle of superposition
Correct Answer: Fermat’s principle
Explanation: Fermat’s principle states that light travels between two points along the path that requires the least time. This principle ensures that rays drawn perpendicular to wavefronts represent the actual optical path.
478. For a point source in 2D, what is the relation between rays and wavefronts?
ⓐ. Rays are radial lines, wavefronts are circular
ⓑ. Rays are parallel, wavefronts are plane
ⓒ. Rays are tangent, wavefronts are elliptical
ⓓ. Rays are random, wavefronts irregular
Correct Answer: Rays are radial lines, wavefronts are circular
Explanation: A point source produces circular wavefronts in 2D. Rays extend radially outward from the center, always perpendicular to the circular wavefronts.
479. In refraction, how do rays and wavefronts behave when light enters from air to glass obliquely?
ⓐ. Wavefront bends, ray remains perpendicular to it
ⓑ. Ray bends, wavefront remains straight
ⓒ. Both bend independently
ⓓ. Neither changes
Correct Answer: Wavefront bends, ray remains perpendicular to it
Explanation: When light enters glass, speed decreases and wavelength changes. The wavefront bends at the interface. Rays, being normal to the wavefronts, also change direction accordingly while remaining perpendicular to the new wavefront.
480. Why are rays often used instead of wavefronts in geometrical optics?
ⓐ. Rays show wave phase directly
ⓑ. Rays are simpler to draw and track, while wavefronts represent phase surfaces
ⓒ. Wavefronts cannot explain reflection
ⓓ. Wavefronts cannot explain refraction
Correct Answer: Rays are simpler to draw and track, while wavefronts represent phase surfaces
Explanation: Wavefronts provide a complete description of wave phase but are harder to draw for complex systems. Rays, always normal to wavefronts, give a simplified model of light propagation used in geometrical optics.
481. A parallel beam of light passes through a convex lens of focal length $20 \, \text{cm}$. Where will the rays converge?
ⓐ. 10 cm from lens
ⓑ. 15 cm from lens
ⓒ. 20 cm from lens
ⓓ. 25 cm from lens
Correct Answer: 20 cm from lens
Explanation: For parallel rays, image forms at the focus.
Here, focal length $f = 20 \, \text{cm}$.
Thus, rays converge at 20 cm from the lens on the principal axis.
482. A concave mirror has focal length $15 \, \text{cm}$. If an object is placed at $30 \, \text{cm}$, where is the image formed?
ⓐ. 10 cm
ⓑ. 15 cm
ⓒ. 20 cm
ⓓ. 30 cm
Correct Answer: 30 cm
Explanation: Formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.
Here, $f = 15 \, \text{cm}, u = 30 \, \text{cm}$.
$\frac{1}{15} = \frac{1}{30} + \frac{1}{v}$.
$\frac{1}{v} = \frac{1}{15} – \frac{1}{30} = \frac{2 – 1}{30} = \frac{1}{30}$.
So, $v = 30 \, \text{cm}$.
483. A light ray is incident at $30^\circ$ on a glass slab with refractive index $1.5$. Find angle of refraction.
ⓐ. $1.95^\circ$
ⓑ. $20^\circ$
ⓒ. $22.50^\circ$
ⓓ. $19.5^\circ$
Correct Answer: $19.5^\circ$
Explanation: Snell’s law: $n_1 \sin i = n_2 \sin r$.
Here, $n_1 = 1, i = 30^\circ, n_2 = 1.5$.
$\sin r = \frac{\sin 30^\circ}{1.5} = \frac{0.5}{1.5} = 0.333$.
$r = \arcsin(0.333) \approx 19.5^\circ$.
484. A convex lens forms a real image at 40 cm for an object placed at 60 cm. Find focal length.
ⓐ. 12 cm
ⓑ. 15 cm
ⓒ. 24 cm
ⓓ. 30 cm
Correct Answer: 24 cm
Explanation: Lens formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Here, $u = -60, v = +40$.
$\frac{1}{f} = \frac{1}{40} + \frac{1}{60}$.
$\frac{1}{f} = \frac{3 + 2}{120} = \frac{5}{120} = \frac{1}{24}$.
So, $f = 24 \, \text{cm}$.
485. A plane mirror produces an image of a point object at 2 m behind it. What is the object distance?
ⓐ. 1 m
ⓑ. 2 m
ⓒ. 3 m
ⓓ. 4 m
Correct Answer: 2 m
Explanation: In plane mirrors, image distance = object distance. Hence, object must also be 2 m in front of the mirror.
486. In acoustics, why are whispering galleries curved?
ⓐ. To amplify sound by resonance
ⓑ. To focus sound waves by reflection along curved surfaces
ⓒ. To eliminate interference
ⓓ. To reduce diffraction
Correct Answer: To focus sound waves by reflection along curved surfaces
Explanation: In whispering galleries, curved geometry reflects sound waves along the walls. Rays follow the surface and meet at distant points, allowing whispers to be heard clearly across the gallery.
487. A sound wave of frequency 200 Hz travels in air with speed 340 m/s. Find its wavelength.
ⓐ. 1.4 m
ⓑ. 1.5 m
ⓒ. 1.6 m
ⓓ. 1.7 m
Correct Answer: 1.7 m
Explanation: Formula: $\lambda = v/f$.
Here, $v = 340, f = 200$.
$\lambda = \frac{340}{200} = 1.7 \, \text{m}$.
488. A tuning fork of 512 Hz produces resonance in an air column of length 16.5 cm closed at one end. Find speed of sound in air.
ⓐ. 320 m/s
ⓑ. 330 m/s
ⓒ. 340 m/s
ⓓ. 338 m/s
Correct Answer: 338 m/s
Explanation: For closed pipe: $L = \frac{\lambda}{4}$.
So, $\lambda = 4L = 4 \times 0.165 = 0.66 \, \text{m}$.
$v = f \lambda = 512 \times 0.66 \approx 338.0 \, \text{m/s}$.
489. In a hall, sound intensity reduces due to multiple reflections and absorption. This property is called:
ⓐ. Diffraction
ⓑ. Reverberation
ⓒ. Dispersion
ⓓ. Interference
Correct Answer: Reverberation
Explanation: Reverberation is persistence of sound due to multiple reflections in a hall. Excess reverberation causes poor audibility, while controlled reverberation enhances richness of sound in auditoriums.
490. A hall requires reverberation time of 1.5 s. If its volume is $3000 \, \text{m}^3$, find total absorption required using Sabine’s formula $T = \frac{0.161 V}{A}$.
ⓐ. 250 m$^2$
ⓑ. 300 m$^2$
ⓒ. 322 m$^2$
ⓓ. 350 m$^2$
Correct Answer: 322 m$^2$
Explanation: Formula: $T = 0.161 \frac{V}{A}$.
Here, $T = 1.5, V = 3000$.
$A = \frac{0.161 \times 3000}{1.5}$.
$A = \frac{483}{1.5}$.
$A = 322 \, \text{m}^2$.
491. Which of the following is the general differential equation of a wave motion in one dimension?
ⓐ. $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$
ⓑ. $\frac{\partial y}{\partial t} = v \frac{\partial y}{\partial x}$
ⓒ. $\frac{\partial y}{\partial t} = \frac{\partial y}{\partial x}$
ⓓ. $\frac{\partial^2 y}{\partial x^2} = \frac{\partial y}{\partial t}$
Correct Answer: $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$
Explanation: The one-dimensional wave equation relates the second time derivative of displacement to its second spatial derivative, multiplied by $v^2$, where $v$ is the wave speed. This is the fundamental differential equation describing wave motion.
492. The wave equation $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$ is of which type?
ⓐ. Ordinary differential equation
ⓑ. First-order partial differential equation
ⓒ. Second-order partial differential equation
ⓓ. Third-order partial differential equation
Correct Answer: Second-order partial differential equation
Explanation: The wave equation involves second derivatives with respect to both time and position. Since more than one independent variable is present ($x, t$), it is a second-order partial differential equation.
493. Which function is a solution of the one-dimensional wave equation?
ⓐ. $y(x,t) = A \sin(kx – \omega t)$
ⓑ. $y(x,t) = A e^{kx}$
ⓒ. $y(x,t) = A t^2 + Bx$
ⓓ. $y(x,t) = A x t$
Correct Answer: $y(x,t) = A \sin(kx – \omega t)$
Explanation: Substituting $y = A \sin(kx – \omega t)$ into the wave equation gives $\omega^2 = v^2 k^2$, which is satisfied for harmonic waves. Other options do not satisfy the second-order relation.
494. A stretched string obeys the wave equation. If displacement is given by $y(x,t) = A \cos(kx – \omega t)$, what relation must hold between $\omega$, $k$, and $v$?
ⓐ. $\omega = v k$
ⓑ. $\omega = v/k$
ⓒ. $\omega^2 = v/k$
ⓓ. $\omega = v^2 k$
Correct Answer: $\omega = v k$
Explanation: For sinusoidal wave solutions, the dispersion relation is $\omega^2 = v^2 k^2$. Taking positive roots, we get $\omega = v k$.
495. A string of length $1 \, \text{m}$ has wave speed $100 \, \text{m/s}$. What is the fundamental frequency?
ⓐ. 25 Hz
ⓑ. 50 Hz
ⓒ. 100 Hz
ⓓ. 200 Hz
Correct Answer: 50 Hz
Explanation: Formula: $f = \frac{v}{2L}$.
Here, $v = 100, L = 1$.
$f = \frac{100}{2 \times 1} = 50 \, \text{Hz}$.
496. The general solution of the wave equation is written as:
ⓐ. $y(x,t) = f(x+vt)$
ⓑ. $y(x,t) = f(x-vt)$
ⓒ. $y(x,t) = f(x+vt) + g(x-vt)$
ⓓ. $y(x,t) = f(x-t)$
Correct Answer: $y(x,t) = f(x+vt) + g(x-vt)$
Explanation: The general solution is a superposition of two functions: one representing a wave traveling in the positive x-direction, the other in the negative x-direction, both moving with speed $v$.
497. A wave is described by $y(x,t) = 0.05 \sin(4\pi t – 0.02\pi x)$. Find wave speed.
ⓐ. 100 m/s
ⓑ. 250 m/s
ⓒ. 400 m/s
ⓓ. 200 m/s
Correct Answer: 200 m/s
Explanation: Here, $\omega = 4\pi \, \text{rad/s}, k = 0.02\pi \, \text{rad/m}$.
Wave speed $v = \omega / k$.
$v = \frac{4\pi}{0.02\pi} = 200$.
Correction: Actually $\frac{4}{0.02} = 200$.
498. For a wave equation, if wave speed $v = 300 \, \text{m/s}$, frequency $f = 150 \, \text{Hz}$, what is wavelength?
ⓐ. 3.5 m
ⓑ. 3.0 m
ⓒ. 2.5 m
ⓓ. 2.0 m
Correct Answer: 2.0 m
Explanation: Formula: $\lambda = v/f$.
Here, $v = 300, f = 150$.
$\lambda = 300/150 = 2.0 \, \text{m}$.
499. Which physical situation can be modeled using the wave equation?
ⓐ. Oscillation of a pendulum
ⓑ. Heat conduction in a rod
ⓒ. Vibrations of a stretched string
ⓓ. Exponential decay of current
Correct Answer: Vibrations of a stretched string
Explanation: Vibrations in a string are governed by the wave equation since they involve propagation of transverse waves with finite speed. Heat conduction obeys diffusion equation, not wave equation.
500. A wave equation is given by $\frac{\partial^2 y}{\partial t^2} = 900 \frac{\partial^2 y}{\partial x^2}$. What is the wave speed?
ⓐ. 20 m/s
ⓑ. 25 m/s
ⓒ. 30 m/s
ⓓ. 35 m/s
Correct Answer: 30 m/s
Explanation: Here, coefficient of $\frac{\partial^2 y}{\partial x^2}$ is $v^2$.
So, $v^2 = 900$.
Hence, $v = \sqrt{900} = 30 \, \text{m/s}$.
You are now on Class 11 Physics MCQs – Chapter 15: Waves (Part 5). This section continues with advanced concepts as per the NCERT/CBSE syllabus including Fourier analysis of waves, wave packet dynamics, and differential equations of wave motion. These higher-order topics are challenging but extremely useful for aspirants of JEE, NEET, and competitive exams. Along with theoretical clarity, this part also focuses on calculation-based numerical questions that are commonly seen in both board exams and entrance tests. From the total of 550 MCQs with answers and explanations, Part 5 contains another 100 questions to help complete your systematic preparation.
- 👉 Total MCQs in this chapter: 550.
- 👉 This page contains: Fifth set of 100 solved MCQs.
- 👉 Highly recommended for JEE/NEET aspirants and board exam students.
- 👉 For more chapters, subjects, or classes, check the navigation bar above.
- 👉 To complete the chapter, open the Part 6 button above for the final 50 questions.