Optics Questions and Answers

Answer: C

Light travels through the core by total internal reflection.

An optical fibre has a core of higher refractive index than its cladding. Light that strikes the core-cladding boundary at more than the critical angle is reflected completely back into the core. It zigzags along the fibre, even round bends, with very little loss. Endoscopes use bundles of such fibres to carry light in and the image out.

Refraction is what happens to light that crosses the boundary below the critical angle, and that light is lost. Diffraction and scattering are loss mechanisms, not the transmission principle.

Answer: A

The refractive index is the ratio of the velocity of light in vacuum to its velocity in the medium.

\[ n = \frac{c}{v} \]

Light slows down in a material, so n is greater than 1: about 1.33 for water and 1.5 for glass.

The inverted ratio, v/c, would be less than 1 for every material. The product of wavelength and frequency is the speed of light in the medium, not its refractive index. For light going from the medium into air the critical angle obeys sin C = 1/n, so n is the reciprocal of sin C, not of the angle.

Answer: B

The power is +5 D.

Power in dioptres is the reciprocal of focal length in metres, and a convex (converging) lens has positive power:

\[ P = \frac{1}{f} = \frac{1}{0.20\ \text{m}} = +5\ \text{D} \]

+2 D would need a focal length of 50 cm. The negative values belong to a concave (diverging) lens. -5 D is the right size but the wrong sign, which is the slip of forgetting that a convex lens converges.

Answer: C

The numerical aperture determines the fibre's light-gathering ability.

NA is the sine of the largest angle at which light entering the fibre will still be guided by total internal reflection:

\[ \text{NA} = \sin\theta_{max} = \sqrt{n_1^{2} - n_2^{2}} \]

A larger NA means a wider acceptance cone, so the fibre collects more light.

NA is set by the core and cladding indices, not by the fibre's diameter. It does not fix the power output, which depends on the source and losses, or the wavelength of light carried.

Answer: A

The He-Ne laser commonly emits at 632.8 nm.

The familiar red beam of a helium-neon laser comes from a transition in neon atoms, which helium atoms excite by collision. It is used for alignment, barcode scanners and low-level laser therapy.

The other wavelengths belong to other lasers. 488 nm is a blue line of the argon-ion laser, 1064 nm is the infrared output of the Nd:YAG laser, and 532 nm is the green light from frequency-doubled Nd:YAG.

Answer: A

Its focal length is Increased.

By the lens-maker's formula, the power of a lens depends on how much its refractive index exceeds that of the medium around it:

\[ \begin{aligned} \frac{1}{f} &= \left(\frac{n_{g}}{n_{m}} - 1\right) \\ &\quad \times \left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \end{aligned} \]

For glass (1.5) in air the bracket is 0.5; in water (1.33) it is only about 0.125. The lens bends light less, so its focal length becomes about four times longer. It does not decrease, stay the same or become zero.

Answer: C

The right lens is a convex lens of focal length 5cm.

A convex lens used as a simple magnifier gives an enlarged, upright image. Its magnifying power is greater when the focal length is shorter, roughly \(1 + D/f\) with \(D = 25\) cm:

\[ \begin{aligned} f = 5\ \text{cm} &: \ 1 + 25/5 = 6 \\ f = 50\ \text{cm} &: \ 1 + 25/50 = 1.5 \end{aligned} \]

So the 5 cm convex lens magnifies far more than the 50 cm one. Concave lenses always form smaller images, so they cannot magnify small print at all.