101. Total internal reflection can occur only when light tries to travel
ⓐ. from a rarer medium to a denser medium
ⓑ. from a denser medium to a rarer medium
ⓒ. normally from one medium into another medium
ⓓ. along the boundary between two identical media
Correct Answer: from a denser medium to a rarer medium
Explanation: Total internal reflection is possible only when light is incident from an optically denser medium into an optically rarer medium. In that case, the refracted ray bends away from the normal as the angle of incidence increases. At a particular angle, the refracted ray just grazes the boundary, giving \(r=90^\circ\). Beyond that angle, no refracted wavefront is formed in the rarer medium and the light is reflected back into the denser medium. If light travels from rarer to denser medium, it bends toward the normal and the limiting grazing refracted ray condition is not reached.
102. Inside glass, a ray approaches a glass-air boundary. As the angle of incidence in glass is gradually increased, the angle of refraction in air
ⓐ. reaches \(90^\circ\) at the critical angle
ⓑ. decreases to \(0^\circ\) at the critical angle
ⓒ. stays equal to the incidence angle
ⓓ. becomes independent of incidence angle
Correct Answer: reaches \(90^\circ\) at the critical angle
Explanation: Glass is optically denser than air, so light going from glass to air bends away from the normal. As the angle of incidence in glass increases, the refracted angle in air also increases. At the critical angle \(C\), the refracted ray just travels along the boundary, so \(r=90^\circ\). This limiting case separates ordinary refraction from total internal reflection. Beyond \(C\), the refracted ray in air is not obtained because the refracted wavefront cannot be constructed as a propagating wave in the second medium.
103. For light going from medium \(1\) to medium \(2\), where \(n_1\gt n_2\), the critical angle \(C\) satisfies
ⓐ. \(\sin C=\frac{n_1}{n_2}\)
ⓑ. \(\tan C=\frac{n_2}{n_1}\)
ⓒ. \(\sin C=\frac{n_2}{n_1}\)
ⓓ. \(\cos C=\frac{n_1}{n_2}\)
Correct Answer: \(\sin C=\frac{n_2}{n_1}\)
Explanation: The critical angle is defined for light travelling from a denser medium into a rarer medium. At the critical angle, the angle of refraction becomes \(90^\circ\). Using Snell's law,
\[
n_1\sin C=n_2\sin90^\circ
\]
Since \(\sin90^\circ=1\), this becomes
\[
n_1\sin C=n_2
\]
Therefore,
\[
\sin C=\frac{n_2}{n_1}
\]
This ratio is less than \(1\) only when \(n_1\gt n_2\), which is exactly the condition for total internal reflection.
104. Light travels from a medium of refractive index \(1.5\) into air of refractive index \(1.0\). The critical angle \(C\) is such that
ⓐ. \(\sin C=\frac{3}{2}\)
ⓑ. \(\sin C=\frac{1}{2}\)
ⓒ. \(\sin C=1\)
ⓓ. \(\sin C=\frac{2}{3}\)
Correct Answer: \(\sin C=\frac{2}{3}\)
Explanation: \( \textbf{Given:} \) The incident medium has \(n_1=1.5\), and air has \(n_2=1.0\).
\( \textbf{Required:} \) The value of \(\sin C\).
\( \textbf{Condition:} \) Critical angle is possible because light is going from larger \(n\) to smaller \(n\).
\( \textbf{Critical-angle relation:} \)
\[
\sin C=\frac{n_2}{n_1}
\]
\( \textbf{Substitution:} \)
\[
\sin C=\frac{1.0}{1.5}
\]
\( \textbf{Simplification:} \)
\[
\sin C=\frac{10}{15}=\frac{2}{3}
\]
\( \textbf{Range check:} \) \(\frac{2}{3}\) is less than \(1\), so it is physically possible as a sine value.
\( \textbf{Final answer:} \) The critical-angle condition is \(\sin C=\frac{2}{3}\).
105. A table describes rays incident from a denser medium into a rarer medium.
| Row | Angle of incidence \(i\) | Expected result |
|---|
| P | \(i\lt C\) | Refraction occurs into the rarer medium |
| Q | \(i=C\) | Refracted ray grazes the boundary |
| R | \(i\gt C\) | Total internal reflection occurs |
| S | \(i\gt C\) | Refracted ray bends toward the normal in the rarer medium |
The row that does not fit total internal reflection is
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row P
ⓓ. Row S
Correct Answer: Row S
Explanation: For \(i\lt C\), the ray refracts into the rarer medium and bends away from the normal. At \(i=C\), the refracted ray just grazes the interface, so \(r=90^\circ\). For \(i\gt C\), no refracted ray is formed in the rarer medium and the ray is totally internally reflected. Row S is wrong because it describes ordinary refraction after the critical angle has been exceeded. The phrase \(i\gt C\) should immediately suggest reflection-only behaviour in this denser-to-rarer situation.
106. Use the wavefront description below.
A plane wavefront in a denser medium is incident on a boundary with a rarer medium. As the incidence angle increases, the refracted wavefront in the rarer medium becomes more tilted. At one limiting angle, the refracted ray would travel along the boundary.
The limiting angle described in the passage is
ⓐ. angle of emergence through a prism
ⓑ. angle of reflection
ⓒ. critical angle
ⓓ. Brewster angle
Correct Answer: critical angle
Explanation: The passage describes light travelling from a denser medium into a rarer medium. In such a case, the refracted ray bends away from the normal. As the incidence angle increases, the refracted angle also increases. The limiting case is reached when the refracted ray becomes parallel to the boundary, so \(r=90^\circ\). That angle of incidence in the denser medium is called the critical angle. Beyond it, Huygens' construction gives no propagating refracted wavefront in the rarer medium.
107. Assertion: Total internal reflection cannot occur when light travels from air into glass.
Reason: For air to glass, the refracted ray bends toward the normal rather than reaching \(r=90^\circ\) as the limiting refracted case.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, but Reason does not explain Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Air is optically rarer than glass. When light enters glass from air, its speed decreases and the ray bends toward the normal. The refracted angle is then smaller than the incident angle, not larger. Total internal reflection needs the opposite situation: denser-to-rarer travel, where the refracted ray bends away from the normal and can reach \(90^\circ\). The Reason correctly explains why the air-to-glass case cannot produce total internal reflection at the boundary.
108. Inside a liquid of refractive index \(1.25\), light is incident on the liquid-air surface. Taking air as \(n=1.00\), the ray is totally internally reflected when
ⓐ. \(i\lt \sin^{-1}(0.80)\)
ⓑ. \(i=0^\circ\)
ⓒ. \(i=\sin^{-1}(0.80)\)
ⓓ. \(i\gt \sin^{-1}(0.80)\)
Correct Answer: \(i\gt \sin^{-1}(0.80)\)
Explanation: \( \textbf{Given:} \) \(n_1=1.25\) for the liquid and \(n_2=1.00\) for air.
\( \textbf{Critical-angle relation:} \)
\[
\sin C=\frac{n_2}{n_1}
\]
\( \textbf{Substitution:} \)
\[
\sin C=\frac{1.00}{1.25}=0.80
\]
\( \textbf{Critical angle:} \)
\[
C=\sin^{-1}(0.80)
\]
\( \textbf{Condition for total internal reflection:} \) It occurs only when the angle of incidence in the denser medium is greater than the critical angle.
\[
i\gt C
\]
\( \textbf{Therefore:} \)
\[
i\gt \sin^{-1}(0.80)
\]
\( \textbf{At equality:} \) The refracted ray would graze the surface, so reflection-only behaviour starts only beyond that angle.
\( \textbf{Final answer:} \) The condition is \(i\gt\sin^{-1}(0.80)\).
109. A wavefront explanation of total internal reflection says that beyond the critical angle
ⓐ. the reflected wavefront disappears
ⓑ. no forward refracted wavefront is constructed in the rarer medium
ⓒ. a refracted wavefront in the rarer medium carries all the incident energy forward
ⓓ. the frequency of the incident light becomes zero
Correct Answer: no forward refracted wavefront is constructed in the rarer medium
Explanation: In Huygens' explanation, refraction is represented by a refracted wavefront advancing in the second medium. For denser-to-rarer incidence below the critical angle, such a wavefront can be constructed and its normal gives the refracted ray. At the critical angle, the refracted ray lies along the boundary. Beyond that angle, a real refracted ray direction in the rarer medium is not obtained. The disturbance is reflected back into the denser medium, so the reflected wavefront carries the propagating light there.
110. For light going from medium \(1\) to medium \(2\), where \(n_1=1.6\) and \(n_2=1.2\), the angle of incidence is \(60^\circ\). Since \(\sin C=\frac{n_2}{n_1}\), the ray
ⓐ. travels along the boundary because \(i=C\)
ⓑ. refracts toward the normal because \(n_2\lt n_1\)
ⓒ. undergoes total internal reflection
ⓓ. refracts normally into medium \(2\)
Correct Answer: undergoes total internal reflection
Explanation: \( \textbf{Given:} \) \(n_1=1.6\), \(n_2=1.2\), and \(i=60^\circ\).
\( \textbf{Medium condition:} \) Since \(n_1\gt n_2\), the ray is travelling from denser to rarer medium.
\( \textbf{Critical-angle relation:} \)
\[
\sin C=\frac{n_2}{n_1}
\]
\( \textbf{Substitution:} \)
\[
\sin C=\frac{1.2}{1.6}=0.75
\]
\( \textbf{Compare with incidence:} \)
\[
\sin60^\circ=\frac{\sqrt{3}}{2}\approx0.866
\]
\( \textbf{Since sine increases for angles below \(90^\circ\):} \) \(\sin60^\circ\gt \sin C\), so \(60^\circ\gt C\).
\( \textbf{TIR condition:} \)
\[
i\gt C
\]
\( \textbf{Final answer:} \) The ray undergoes total internal reflection.
111. The superposition principle for light waves states that when two waves overlap at a point, the resultant disturbance is
ⓐ. always the larger disturbance alone
ⓑ. always zero at the overlap point
ⓒ. the vector sum of individual disturbances
ⓓ. independent of phase at the overlap point
Correct Answer: the vector sum of individual disturbances
Explanation: Superposition applies when waves overlap in a region where the medium or field response is linear. At a point, the resultant disturbance is obtained by adding the disturbances due to the individual waves at that same point and same instant. Depending on their relative phase, the addition may produce a larger, smaller, or even zero resultant amplitude. It is not always the larger disturbance alone, and it is not always zero. The phase relationship decides whether the waves reinforce or cancel.
112. Two light waves reach the same point with equal amplitudes and the same phase. The resultant amplitude at that point is
ⓐ. half of one amplitude
ⓑ. twice one amplitude
ⓒ. zero
ⓓ. equal to one amplitude
Correct Answer: twice one amplitude
Explanation: When two waves are in the same phase, their displacements at a point are in the same direction at the same instant. If each wave has amplitude \(a\), the maximum resultant displacement becomes \(a+a=2a\). This is constructive superposition at the amplitude level. The intensity is then related to the square of the resultant amplitude, so the intensity change is not simply a doubling. The amplitude result must be found first before discussing intensity.
113. Two equal-amplitude light waves arrive at a point exactly out of phase. The resultant disturbance is zero at that point because
ⓐ. the waves have different frequencies
ⓑ. the two waves arrive with the same displacement sign
ⓒ. light-wave amplitudes cannot be added
ⓓ. equal opposite displacements cancel
Correct Answer: equal opposite displacements cancel
Explanation: If two equal-amplitude waves are exactly out of phase, one wave has a positive displacement when the other has an equal negative displacement at the point. Superposition then gives a resultant displacement of zero. This is the amplitude-level basis of destructive interference. The waves do not need to have different frequencies for this cancellation; stable cancellation is actually easiest to observe when the waves have the same frequency and a fixed phase relation. The cancellation is local and depends on the phase at that point.
114. A record of overlapping waves is shown below.
| Row | Condition at a point | Result by superposition |
|---|
| P | Equal amplitudes, same phase | Resultant amplitude \(2a\) |
| Q | Equal amplitudes, opposite phase | Resultant amplitude \(0\) |
| R | Unequal amplitudes, opposite phase | Resultant amplitude equals difference of amplitudes |
| S | Same phase | Resultant amplitude is always zero |
The row that needs correction is
ⓐ. Row S
ⓑ. Row P
ⓒ. Row R
ⓓ. Row Q
Correct Answer: Row S
Explanation: Row P correctly describes constructive addition of two equal amplitudes in the same phase. Row Q correctly gives complete cancellation for equal amplitudes in opposite phase. Row R is also acceptable because opposite-phase unequal amplitudes subtract, leaving the difference as the resultant amplitude. Row S is wrong because same-phase waves reinforce rather than cancel. The sign of displacement at the same instant is the deciding feature in superposition.
115. Two coherent light waves have displacements \(y_1=3a\) and \(y_2=2a\) at the same point and same instant. If the displacements are along the same direction, the resultant displacement is
ⓐ. \(6a\)
ⓑ. \(5a\)
ⓒ. \(a\)
ⓓ. \(2a\)
Correct Answer: \(5a\)
Explanation: \( \textbf{Given:} \) \(y_1=3a\) and \(y_2=2a\).
\( \textbf{Condition:} \) The displacements are along the same direction at the same point and same instant.
\( \textbf{Superposition rule:} \)
\[
y=y_1+y_2
\]
\( \textbf{Substitution:} \)
\[
y=3a+2a
\]
\( \textbf{Calculation:} \)
\[
y=5a
\]
\( \textbf{Why not \(a\):} \) The value \(a\) would come from subtracting the displacements, which is not the given direction condition.
\( \textbf{Why not \(6a\):} \) Multiplication of displacements is not the superposition rule.
\( \textbf{Final answer:} \) The resultant displacement is \(5a\).
116. Two overlapping waves at a point have displacements \(4a\) and \(-a\) at the same instant. The resultant displacement is
ⓐ. \(5a\)
ⓑ. \(4a\)
ⓒ. \(3a\)
ⓓ. \(-5a\)
Correct Answer: \(3a\)
Explanation: \( \textbf{Given:} \) The instantaneous displacements are \(4a\) and \(-a\).
\( \textbf{Required:} \) Resultant displacement at the same point and same instant.
\( \textbf{Superposition principle:} \)
\[
y=y_1+y_2
\]
\( \textbf{Substitution:} \)
\[
y=4a+(-a)
\]
\( \textbf{Simplification:} \)
\[
y=3a
\]
\( \textbf{Sign meaning:} \) The negative sign means the second displacement is opposite to the positive direction, so it reduces the resultant.
\( \textbf{Wrong-option check:} \) \(5a\) would ignore the negative sign and add magnitudes only.
\( \textbf{Final answer:} \) The resultant displacement is \(3a\).
117. Intensity in wave optics is connected more directly with
ⓐ. the resultant amplitude squared
ⓑ. the square of path difference only
ⓒ. the reciprocal of wave frequency
ⓓ. the sum of the two wavelengths
Correct Answer: the resultant amplitude squared
Explanation: The observable brightness in interference is connected with intensity, not just displacement. For light waves under the usual conditions, intensity is proportional to the square of the resultant amplitude. This is why doubling the resultant amplitude produces four times the intensity. Superposition is first applied to amplitudes or disturbances, and intensity is then inferred from the square of the resultant amplitude. Adding wavelengths or using only the sign of path difference does not give the brightness.
118. A graph description is given below.
At a fixed point, the resultant amplitude \(A\) of two equal light waves is varied by changing their relative phase. The vertical axis represents intensity \(I\), and the horizontal axis represents resultant amplitude \(A\).
The graph should show that
ⓐ. \(I\propto A^2\)
ⓑ. \(I\propto A\)
ⓒ. \(I\) is independent of \(A\)
ⓓ. \(I\propto \frac{1}{A}\)
Correct Answer: \(I\propto A^2\)
Explanation: The intensity of a light wave is proportional to the square of its amplitude. If the resultant amplitude after superposition is \(A\), then \(I\propto A^2\). This means the graph of \(I\) against \(A\) is not a straight line through the origin with constant slope; it rises more rapidly as \(A\) increases. A doubled resultant amplitude gives four times the intensity. This square dependence is the reason interference brightness changes strongly when the phase relation changes.
119. Consider the following statements about superposition of light waves.
Statement I: The waves must be considered at the same point and the same instant.
Statement II: Resultant displacement is found before resultant intensity is inferred.
Statement III: Two waves always produce a brighter region wherever they overlap.
ⓐ. I, II and III
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I and III only
Correct Answer: I and II only
Explanation: Statement I is true because superposition adds disturbances that exist together at the same point and at the same instant. Statement II is also true because the wave disturbance or amplitude is combined first, and intensity follows from the square of the resultant amplitude. Statement III is false because overlapping waves can produce reinforcement or cancellation depending on phase. Two waves may produce a bright region, a dark region, or an intermediate intensity. Overlap alone is not enough; the relative phase determines the result.
120. A claim says, “If two light waves overlap, energy is created at bright regions and destroyed at dark regions.” The better interpretation is that
ⓐ. energy is destroyed wherever amplitudes cancel
ⓑ. energy is created only at the central bright region
ⓒ. superposition violates conservation of energy
ⓓ. energy is redistributed in the interference pattern
Correct Answer: energy is redistributed in the interference pattern
Explanation: Superposition can make the resultant amplitude large at some points and small or zero at other points. Since intensity depends on the square of resultant amplitude, bright and dark regions appear in the pattern. This does not mean energy is created at bright regions or destroyed at dark regions. The wave energy is redistributed across the pattern, with reinforcement in some regions and cancellation in others. A dark point represents local destructive superposition, not a failure of energy conservation.
