Optics Questions and Answers

Answer: D

The fringe width becomes 0.6 mm.

In Young's experiment the fringe width is \(\beta = \frac{\lambda D}{d}\), so with the same slits and screen \(\beta \propto \lambda\).

\[ \beta' = 0.5 \times \frac{600}{500} = 0.6\ \text{mm} \]

Longer wavelength gives wider fringes, so the answers below 0.5 mm are impossible.

Answer: C

Polaroids are used to control the intensity of light from the objects.

Light reflected from water, roads and glass is partly polarised in the horizontal plane. Polaroid sunglasses, with their transmission axis vertical, block much of this polarised glare while passing the rest, so they reduce the intensity of light reaching the eyes.

They do not change colour, magnify objects or exist for fashion or cost; a polarising filter transmits all colours alike and has no focusing power.

Answer: A

The effective power is 9 D.

Power in dioptres is the reciprocal of focal length in metres. For lenses in contact, the powers add:

\[ \begin{aligned} P_{1} &= \frac{1}{0.20\ \text{m}} = 5\ \text{D} \\ P &= P_{1} + P_{2} = 5 + 4 = 9\ \text{D} \end{aligned} \]

5 D is the first lens alone. The focal length must be in metres, not centimetres, or the power comes out a hundred times too small. 6, 10 and 11 D do not follow from these values.

Answer: D

The false match is concave mirror : interference.

A concave mirror forms images by reflection; it has nothing to do with interference, which is the superposition of coherent waves, as in Young's double-slit experiment or the colours of thin films.

The other pairs are correct: optical fibres carry light by repeated total internal reflection, thin plastic (Polaroid) sheets polarise light, a glass prism disperses white light into its colours, and a glass slab refracts (and laterally displaces) a ray passing through it.

Answer: B

The refractive index is \(\sqrt{2}\).

\[ \begin{aligned} \mu &= \frac{\sin\frac{A + D_{m}}{2}}{\sin\frac{A}{2}} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} \\ &= \frac{1/\sqrt{2}}{1/2} = \sqrt{2} \end{aligned} \]

A refractive index is greater than 1 for glass, which rules out \(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{3}}\) and \(\frac{\sqrt{3}}{2}\); \(\sqrt{3}\) would need a minimum deviation of 60°.

Answer: D

The slit separation is 0.4 mm.

Fringe width \(\beta = \frac{\lambda D}{d}\), so a change in wavelength changes \(\beta\) by \(\frac{\Delta\lambda D}{d}\):

\[ \begin{aligned} d &= \frac{\Delta\lambda D}{\Delta\beta} = \frac{100 \times 10^{-9} \times 1}{0.25 \times 10^{-3}} \\ &= 4 \times 10^{-4}\ \text{m} = 0.4\ \text{mm} \end{aligned} \]

Check: \(\beta\) is 1.25 mm at 500 nm and 1.5 mm at 600 nm. 4 is the same number left in units of \(10^{-4}\) m, and using 600 nm alone instead of the change gives 2.4 mm.

Answer: C

The power of the combination is halved.

For thin lenses in contact, powers add: \(P = \frac{1}{f_{1}} + \frac{1}{f_{2}}\). Doubling each focal length halves each term:

\[ P' = \frac{1}{2f_{1}} + \frac{1}{2f_{2}} = \frac{P}{2} \]

Power is the reciprocal of focal length, so longer focal lengths mean weaker lenses; 'is doubled' reverses that relationship. The result holds whatever the two focal lengths are.

Answer: D

The net power is 25 D.

The power of a lens is \(P = \frac{1}{f}\) with \(f\) in metres, and the powers of thin lenses in contact add.

\[ \begin{aligned} P_1 &= \frac{1}{0.20} = 5 \text{ D} \\ P &= 5 \times 5 = 25 \text{ D} \end{aligned} \]

5 D is the power of one lens, and 20 D comes from treating the focal length in centimetres as a power. The combination's focal length is 4 cm.

Answer: C

The fringe width is 0.5 mm.

Fringe width is \(\beta = \frac{\lambda D}{d}\), with D the distance to the screen and d the slit separation.

\[ \begin{aligned} \beta &= \frac{500 \times 10^{-9} \times 2}{2 \times 10^{-3}} \\ &= 5 \times 10^{-4} \text{ m} = 0.5 \text{ mm} \end{aligned} \]

5 mm is ten times too large, a slip in the powers of ten; keep every length in metres until the last step.

Answer: E

The image is formed at a distance \(\frac{f}{2}\).

Using the mirror formula with the Cartesian sign convention, the object distance is \(u = -f\) and a convex mirror has focal length \(+f\):

\[ \begin{aligned} \frac{1}{v} + \frac{1}{u} &= \frac{1}{f} \\ \frac{1}{v} &= \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \Rightarrow v = \frac{f}{2} \end{aligned} \]

The image is virtual, behind the mirror, halfway between pole and focus. Taking the convex mirror's focal length as negative leads to an infinite or wrong distance; 2f and f are concave-mirror reasoning.