501. A telescope is used to observe a faint star cluster. A larger objective aperture helps primarily by
ⓐ. changing the star colours into a prism spectrum
ⓑ. reducing the focal length of the eyepiece automatically
ⓒ. collecting more light for faint details
ⓓ. making the final image real on a screen for every adjustment
Correct Answer: collecting more light for faint details
Explanation: Faint astronomical objects send very little light to Earth. A larger objective aperture collects more of this light. The image therefore becomes brighter and faint details become easier to observe. The same larger aperture can also help resolution, but its first visible advantage for faint objects is light gathering. It does not automatically change the eyepiece focal length or force every final image to be real.
502. In microscope resolution, using blue light instead of red light can be helpful because blue light has
ⓐ. shorter wavelength
ⓑ. larger wavelength
ⓒ. zero frequency
ⓓ. no diffraction at all
Correct Answer: shorter wavelength
Explanation: Diffraction effects become smaller when wavelength is shorter. Blue light has a shorter wavelength than red light. Therefore, under suitable conditions, blue light can help a microscope resolve finer details than red light. This does not mean diffraction disappears completely. It means the diffraction-limited separation can be reduced when the wavelength is reduced.
503. A simple microscope, a compound microscope, and an astronomical telescope all use lenses, but the object for a telescope is usually
ⓐ. very distant
ⓑ. just inside the focal length of a single magnifier
ⓒ. just beyond the focus of a short objective near the specimen
ⓓ. at the near point \(D\) only
Correct Answer: very distant
Explanation: A telescope is designed to view distant objects such as stars or faraway terrestrial objects. The rays from such objects reach the objective nearly parallel. A simple microscope is used for a nearby small object placed within the focal length of a convex lens. A compound microscope is used for nearby tiny specimens placed just beyond the focus of the objective. The object distance is therefore a major way to distinguish these instruments.
504. The objective in a compound microscope and the objective in an astronomical telescope differ mainly because the microscope objective has
ⓐ. long focal length for distant stars, while the telescope objective has short focal length for specimens
ⓑ. short focal length for magnification; long focal length and aperture for a telescope
ⓒ. negative focal length, while the telescope objective has zero focal length
ⓓ. no image-forming role, while the telescope objective forms all final images alone
Correct Answer: short focal length for magnification; long focal length and aperture for a telescope
Explanation: A compound microscope objective is placed close to a small object and has a very short focal length. This helps it form a large real intermediate image. A telescope objective faces distant objects and usually has a long focal length and large aperture. The long focal length helps angular magnification, while the large aperture collects more light. Both are called objectives, but their design requirements are different.
505. A learner mixes up microscope and telescope formulas. The expression \(M=\frac{L}{f_o}\frac{D}{f_e}\) belongs to
ⓐ. astronomical telescope in normal adjustment
ⓑ. a compound microscope with final image at infinity
ⓒ. a simple microscope with final image at near point
ⓓ. a prism at minimum deviation
Correct Answer: a compound microscope with final image at infinity
Explanation: The expression contains the tube length \(L\), the objective focal length \(f_o\), the eyepiece focal length \(f_e\), and the near point \(D\). This is the approximate magnifying power of a compound microscope when the final image is at infinity. An astronomical telescope in normal adjustment has \(M=\frac{f_o}{f_e}\). A simple microscope uses \(M=\frac{D}{f}\) or \(M=1+\frac{D}{f}\). Prism formulas involve \(A\), \(\delta_m\), and refractive index rather than microscope tube length.
506. A comparison table for optical instruments is given:
| Row | Instrument | First image formed by objective | Usual final image |
|---|
| P | Compound microscope | real, inverted, enlarged | virtual, enlarged |
| Q | Astronomical telescope | real, inverted, diminished | virtual, magnified angularly |
| R | Simple microscope | real image by objective and then eyepiece | always real on screen |
| S | Galilean telescope | uses concave eyepiece | erect final image |
The acceptable rows are
ⓐ. P and R only
ⓑ. Q, R, and S only
ⓒ. P, Q, and S only
ⓓ. P, Q, R, and S
Correct Answer: P, Q, and S only
Explanation: Row P is acceptable because the compound microscope objective forms a real enlarged intermediate image, and the eyepiece gives a final virtual enlarged image. Row Q is acceptable because a telescope objective forms a real diminished image of a distant object, and the eyepiece magnifies it angularly. Row S is acceptable because a Galilean telescope uses a concave eyepiece and gives an erect final image. Row R is not acceptable because a simple microscope is a single convex lens, not an objective-eyepiece system. Its final image is virtual in normal magnifying use.
507. The formula \(M=1+\frac{D}{f}\) and the formula \(M=\frac{f_o}{f_e}\) are used for different instruments. The first applies to a
ⓐ. simple microscope at near point and telescope in normal adjustment
ⓑ. telescope with final image at infinity, and a simple microscope at near point
ⓒ. compound microscope and a prism
ⓓ. reflecting telescope aperture and a glass slab
Correct Answer: simple microscope at near point and telescope in normal adjustment
Explanation: The formula \(M=1+\frac{D}{f}\) contains the near point \(D\) and the focal length of a single convex lens. It describes a simple microscope when the final image is at the near point. The formula \(M=\frac{f_o}{f_e}\) compares objective and eyepiece focal lengths in an astronomical telescope in normal adjustment. The two formulas both describe angular magnification, but the instruments and image conditions are different. Recognising the symbols prevents using a microscope relation in a telescope problem.
508. A compound microscope and an astronomical telescope both have an objective and an eyepiece. The eyepiece in both instruments primarily
ⓐ. forms the first image directly from the object in every case
ⓑ. replaces the retina of the observer
ⓒ. removes the need for refraction
ⓓ. magnifies the intermediate image for the eye
Correct Answer: magnifies the intermediate image for the eye
Explanation: In both a compound microscope and a telescope, the objective first forms an intermediate image. The eyepiece is placed so that this intermediate image can be viewed under a larger angle. It acts like a magnifier for the eye. The objective, not the eyepiece, is responsible for collecting light from the original object and forming the first image. The eyepiece does not replace the retina; the final light still enters the eye and forms an image on the retina.
509. An instrument has a short focal length objective, a short focal length eyepiece, and a tube length much larger than both focal lengths. It is most likely a
ⓐ. astronomical telescope
ⓑ. compound microscope
ⓒ. simple microscope
ⓓ. rectangular glass slab
Correct Answer: compound microscope
Explanation: A compound microscope uses a short focal length objective to form a large real intermediate image. It also uses a short focal length eyepiece to magnify that intermediate image. The tube length is an important part of its magnifying power through the factor \(\frac{L}{f_o}\). A telescope usually has a long focal length objective, not a short one. A simple microscope has only one convex lens and no objective-eyepiece tube arrangement.
510. An instrument has a long focal length objective of large aperture and a short focal length eyepiece. It is adjusted so that the final image is at infinity. This description best matches
ⓐ. astronomical telescope in normal adjustment
ⓑ. a simple microscope at near-point adjustment
ⓒ. a rectangular glass slab
ⓓ. a plane mirror periscope
Correct Answer: astronomical telescope in normal adjustment
Explanation: A long focal length objective and a short focal length eyepiece are typical of an astronomical telescope. The large aperture objective helps collect light from distant objects. Normal adjustment means the final image is at infinity, giving relaxed-eye viewing. The corresponding magnifying power is \(\frac{f_o}{f_e}\). A simple microscope uses a single short focal length convex lens, not a long objective and eyepiece pair.
511. A table lists possible design choices:
| Row | Goal | Suitable design choice |
|---|
| P | Higher telescope magnification with fixed objective | smaller \(f_e\) |
| Q | Higher microscope magnification with fixed \(L\) and \(f_e\) | smaller \(f_o\) |
| R | Brighter telescope image of faint object | larger objective aperture |
| S | Higher simple microscope magnification | larger \(f\) |
The suitable rows are
ⓐ. P and S only
ⓑ. Q, R, and S only
ⓒ. P, Q, R, and S
ⓓ. P, Q, and R only
Correct Answer: P, Q, and R only
Explanation: Row P is suitable because \(M=\frac{f_o}{f_e}\) increases when \(f_e\) decreases for a fixed telescope objective. Row Q is suitable because microscope magnification contains the factor \(\frac{L}{f_o}\), so reducing \(f_o\) increases magnification. Row R is suitable because a larger objective aperture collects more light from faint objects. Row S is not suitable because a simple microscope has \(M=\frac{D}{f}\) or \(M=1+\frac{D}{f}\), both of which increase when \(f\) decreases. Different instruments require different focal length choices.
512. A single convex lens is used to see a small stamp clearly with final image at the near point. A two-lens instrument is used to see bacteria on a slide. A long objective with a short eyepiece is used to see the Moon. The three instruments are respectively
ⓐ. astronomical telescope, simple microscope, compound microscope
ⓑ. compound microscope, telescope, simple microscope
ⓒ. simple microscope, compound microscope, astronomical telescope
ⓓ. prism, glass slab, reflecting mirror
Correct Answer: simple microscope, compound microscope, astronomical telescope
Explanation: A single convex lens used for close viewing is a simple microscope or magnifying glass. A two-lens instrument for very small nearby specimens is a compound microscope. A long objective with a short eyepiece used for distant celestial objects is an astronomical telescope. The object distance and lens arrangement identify the instrument. The same word magnification is used in all three cases, but the optical layout is different.
513. A formula-choice note says: use \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) for a thin lens, \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) for a spherical mirror, and \(n_1\sin i=n_2\sin r\) for refraction at a plane boundary. This note is
ⓐ. unsuitable, because all optical devices use the mirror formula
ⓑ. unsuitable, because Snell's law is used only for reflection
ⓒ. suitable only when all distances are positive
ⓓ. suitable, because the formula must match the optical situation
Correct Answer: suitable, because the formula must match the optical situation
Explanation: Different optical situations have different governing relations. A thin lens forms images by refraction at two surfaces and uses \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\). A spherical mirror forms images by reflection and uses \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\). Refraction at a plane boundary is described by Snell's law, \(n_1\sin i=n_2\sin r\). Using the correct formula is as important as substituting the correct numerical values.
514. An object is \(30\,cm\) in front of a convex lens of focal length \(20\,cm\), while another identical object is \(30\,cm\) in front of a concave mirror of focal length \(-20\,cm\). The lens image distance and mirror image distance are respectively
ⓐ. \(-60\,cm\) and \(+60\,cm\)
ⓑ. \(+12\,cm\) and \(-12\,cm\)
ⓒ. \(+60\,cm\) and \(-60\,cm\)
ⓓ. \(-12\,cm\) and \(+12\,cm\)
Correct Answer: \(+60\,cm\) and \(-60\,cm\)
Explanation: \( \textbf{For the convex lens:} \)
\[
f=+20\,cm,\quad u=-30\,cm
\]
Thin lens formula:
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
\[
\frac{1}{20}=\frac{1}{v}+\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}
\]
\[
v=+60\,cm
\]
\( \textbf{For the concave mirror:} \)
\[
f=-20\,cm,\quad u=-30\,cm
\]
Mirror formula:
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
\[
\frac{1}{-20}=\frac{1}{v}+\frac{1}{-30}
\]
\[
\frac{1}{v}=-\frac{1}{20}+\frac{1}{30}=-\frac{1}{60}
\]
\[
v=-60\,cm
\]
\( \textbf{Final answer:} \) \(+60\,cm\) for the lens and \(-60\,cm\) for the mirror.
515. A real object is placed \(30\,cm\) in front of a convex lens of focal length \(+20\,cm\), and another identical object is placed \(30\,cm\) in front of a concave mirror of focal length \(-20\,cm\). The magnifications are respectively
ⓐ. \(-2\) and \(-2\)
ⓑ. \(+2\) and \(+2\)
ⓒ. \(-2\) and \(+2\)
ⓓ. \(+2\) and \(-2\)
Correct Answer: \(-2\) and \(-2\)
Explanation: \( \textbf{For the convex lens:} \)
From the previous image-position result for \(f=+20\,cm\) and \(u=-30\,cm\),
\[
v=+60\,cm
\]
Lens magnification is
\[
m_{\text{lens}}=\frac{v}{u}
\]
\[
m_{\text{lens}}=\frac{+60}{-30}=-2
\]
\( \textbf{For the concave mirror:} \)
From the mirror calculation for \(f=-20\,cm\) and \(u=-30\,cm\),
\[
v=-60\,cm
\]
Mirror magnification is
\[
m_{\text{mirror}}=-\frac{v}{u}
\]
\[
m_{\text{mirror}}=-\frac{-60}{-30}=-2
\]
\( \textbf{Final answer:} \) both magnifications are \(-2\), though the image distances have opposite signs because lens and mirror sign meanings differ.
516. A learner has to select a formula for each situation:
| Row | Situation | Suitable formula |
|---|
| P | Thin lens image formation | \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) |
| Q | Spherical mirror image formation | \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) |
| R | Plane-boundary refraction | \(n_1\sin i=n_2\sin r\) |
| S | Prism at minimum deviation | \(n=\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\) |
The acceptable rows are
ⓐ. P and Q only
ⓑ. P, Q, R, and S
ⓒ. Q and R only
ⓓ. P, R, and S only
Correct Answer: P, Q, R, and S
Explanation: Row P is suitable because the thin lens formula has the form \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) under the usual Cartesian sign convention. Row Q is suitable because the spherical mirror formula uses \(\frac{1}{v}+\frac{1}{u}\). Row R is suitable because refraction at a plane boundary is governed by Snell's law. Row S is suitable because the prism refractive-index formula at minimum deviation uses half-angle sine terms. The same symbols \(u\), \(v\), \(f\), \(n\), and angles cannot be moved between formulas without matching the optical situation.
517. A thin convex lens has power \(+5\,\text{D}\). A real object is placed \(30\,cm\) in front of it. The image distance is
ⓐ. \(-60\,cm\)
ⓑ. \(+12\,cm\)
ⓒ. \(-12\,cm\)
ⓓ. \(+60\,cm\)
Correct Answer: \(+60\,cm\)
Explanation: \( \textbf{Convert power to focal length:} \)
\[
P=+5\,\text{D}
\]
\[
f=\frac{1}{P}=\frac{1}{5}\,m=0.20\,m
\]
\[
f=+20\,cm
\]
The real object is in front of the lens:
\[
u=-30\,cm
\]
Use the thin lens formula:
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
Substitute:
\[
\frac{1}{20}=\frac{1}{v}-\frac{1}{-30}
\]
\[
\frac{1}{20}=\frac{1}{v}+\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}
\]
\[
v=+60\,cm
\]
\( \textbf{Final answer:} \) \(+60\,cm\), so the image is real and formed on the far side of the lens.
518. A concave mirror and a convex lens both have focal length magnitude \(20\,cm\). A real object is kept \(10\,cm\) in front of each. The image nature in both cases is
ⓐ. real, inverted, and enlarged
ⓑ. virtual, erect, and diminished
ⓒ. real, inverted, and diminished
ⓓ. virtual, erect, and enlarged
Correct Answer: virtual, erect, and enlarged
Explanation: For a concave mirror, an object placed between the pole \(P\) and focus \(F\) forms a virtual, erect, enlarged image behind the mirror. Here the object distance magnitude \(10\,cm\) is less than the focal length magnitude \(20\,cm\), so the object is inside the focal length. For a convex lens, an object placed between the optical centre \(O\) and focus \(F_1\) also forms a virtual, erect, enlarged image on the object side. In both devices, the actual reflected or refracted rays diverge after interaction. Their backward extensions meet to form the virtual image.
519. A real object is placed before an optical element. The image is real, inverted, and diminished. This image can be formed by
ⓐ. a concave lens for any real-object position
ⓑ. convex lens beyond \(2F_1\), or concave mirror beyond \(C\)
ⓒ. a convex mirror for any real-object position
ⓓ. a plane mirror or convex mirror for any real-object position
Correct Answer: convex lens beyond \(2F_1\), or concave mirror beyond \(C\)
Explanation: A convex lens forms a real, inverted, diminished image when the object is placed beyond \(2F_1\). A concave mirror forms the same type of image when the object is placed beyond the centre of curvature \(C\). A concave lens gives a virtual, erect, diminished image for a real object. A convex mirror also gives a virtual, erect, diminished image for a real object. A plane mirror forms a virtual, erect image of the same size, so it cannot match the stated image nature.
520. An optical system first forms a real image \(40\,cm\) to the right of a convex lens. A second convex lens is placed \(25\,cm\) to the right of the first lens, before that image is reached. For the second lens, the object distance is
ⓐ. \(-15\,cm\)
ⓑ. \(+40\,cm\)
ⓒ. \(+15\,cm\)
ⓓ. \(-25\,cm\)
Correct Answer: \(+15\,cm\)
Explanation: \( \textbf{Position information:} \)
The first lens would form its image \(40\,cm\) to the right of itself.
The second lens is \(25\,cm\) to the right of the first lens.
The would-be image point is therefore
\[
40\,cm-25\,cm=15\,cm
\]
to the right of the second lens.
Rays reaching the second lens are converging toward this point.
For the second lens, this point acts as a virtual object.
With light travelling left to right, the right side of the second lens is positive.
\[
u_2=+15\,cm
\]
\( \textbf{Final answer:} \) \(+15\,cm\), because the object for the second lens lies on its outgoing side.
