Optics Questions and Answers

Answer: A

The frequency shows No Change.

Frequency is set by the source of the light and does not change when light passes from one medium to another. What changes on going from a denser to a rarer medium is the speed, which increases, and so the wavelength, which increases in the same proportion (v = fλ).

Options saying the frequency increases, decreases, or rises and then falls confuse frequency with speed or wavelength. If the frequency changed, the colour of light would change on entering water or glass, which does not happen.

Answer: D

No image will be formed.

An image forms when reflected rays from each point of the object stay together in an orderly way, as from a smooth mirror (regular reflection). In completely diffused reflection, from a rough surface such as paper or a wall, parallel rays are scattered in all directions, so they do not meet or appear to meet at any point.

The other options describe images that need regular reflection. A virtual image of the same size is what a plane mirror forms. Real enlarged and virtual enlarged images are formed by concave mirrors. A diffusing surface forms none of these.

Answer: D

The properties that explain shadows are D and A.

A shadow forms because light travels in straight lines (A) and cannot pass through an opaque object (D). The object blocks the straight rays, and the region behind it that the rays cannot reach is dark.

The high speed of light (B) and its being composed of seven colours (C) are true facts but play no part in shadow formation, so every option that includes B or C is wrong.

Answer: C

X and Y are Periscope, reflection.

A periscope has two plane mirrors fixed parallel to each other at 45° to the tube. Light from an object is reflected by the top mirror down the tube and by the bottom mirror into the eye, which lets the viewer see over walls or around corners, as in submarines.

A telescope is used to see distant objects, not around corners, and a kaleidoscope makes patterns by multiple reflections. A periscope does not work by dispersion, the splitting of light into colours.

Answer: B

The correct statement is that a concave mirror forms an erect as well as an inverted image.

The image formed by a concave mirror depends on where the object is. When the object is between the pole and the focus, the image is virtual, erect and enlarged, as in a shaving mirror. When the object is beyond the focus, the image is real and inverted.

So a concave mirror does not always form an erect image. A convex mirror always forms a virtual, erect and diminished image, whatever the position of the object, which is why it is used as a rear-view mirror; it never forms an inverted image.

Answer: C

The mirrors are A - plane, B - concave, C - concave, D - convex.

• A: a virtual, erect image of the same size is formed by a plane mirror.
• B: a real, inverted image of the same size is formed by a concave mirror with the object at the centre of curvature.
• C: a real, inverted, enlarged image is formed by a concave mirror with the object between the focus and the centre of curvature.
• D: a virtual, erect, diminished image is always formed by a convex mirror.

A convex mirror never forms a real image, which rules out the options giving B or C as convex.

Answer: C

The fringe width is \(\beta = \lambda D / d\). Placing the apparatus in a liquid of refractive index \(\mu\) shortens the wavelength to \(\lambda / \mu\), so the fringe width falls in the same ratio.

\[ \begin{aligned} \beta' = \frac{\beta}{\mu} = \frac{2.4}{1.2} \\ &= 2.0\ \text{mm} \end{aligned} \]

The wavelength of 500 nm is not needed once the fringe width in air is known. The value 1.0 mm would need a refractive index of 2.4, 4.0 mm would be an increase, and 0.5 mm has no basis. The answer is 2.0 mm, option C.

Answer: C

The fringe width in Young's experiment is \(\beta = \lambda D / d\).

\[ \begin{aligned} \beta \\ &= \frac{500 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} \\ &= 2.5 \times 10^{-4}\ \text{m} = 0.25\ \text{mm} \end{aligned} \]

The slit separation of 3 mm must be converted to metres and the wavelength of 500 nm to \(5 \times 10^{-7}\) m. The values 0.25 cm and 2.5 cm are ten and a hundred times too large, and 0.25 m is a thousand times too large, which would come from mixing up millimetres and metres. The answer is 0.25 mm, option C.

Answer: A

The correct match is P → 5, Q → 4, R → 1, S → 3.

• Aurora Borealis (P → 5): charged particles from the Sun, guided by the Earth's magnetic field into the polar upper atmosphere, excite oxygen and nitrogen atoms, which then emit light (green and red from oxygen, blue and violet from nitrogen).
• Partially polarized sunlight (Q → 4): light scattered by air molecules is polarized, most strongly at 90° to the direction of the Sun.
• Rainbow (R → 1): sunlight is refracted and dispersed on entering a raindrop and reflected from its back surface, so the colours separate.
• Dark and bright fringes (S → 3): fringes arise from the superposition of waves that have spread out from apertures or edges, the diffraction (and interference) of light.

Total internal reflection (2) is not needed for any of the four: the reflection inside a raindrop happens below the critical angle. That rules out every option that uses entry 2.

Answer: B

The smallest minimum deviation is 4.8°.

For a thin prism \(D_{m} = (n - 1)A\), so \(D_{m}\) is smallest where \(n(\lambda)\) is smallest. Set the derivative to zero:

\[ \begin{aligned} \frac{dn}{d\lambda} &= \alpha - \frac{2\beta}{\lambda^{3}} = 0 \\ \lambda^{3} &= \frac{2\beta}{\alpha} = \frac{0.192}{3} = 0.064 \\ \lambda_{\text{min}} &= 0.4\ \mu\text{m} \end{aligned} \]

The second derivative \(6\beta/\lambda^{4}\) is positive, so this is a minimum of \(n\):

\[ \begin{aligned} n &= 3(0.4) + \frac{0.096}{0.16} = 1.2 + 0.6 = 1.8 \\ D_{m} &= (1.8 - 1)(6^{\circ}) = 4.8^{\circ} \end{aligned} \]

The other options would need indices of about 2.07 (6.4°), 1.53 (3.2°) and 1.40 (2.4°) at the minimum, none of which the dispersion relation gives.