Optics Questions and Answers

Answer: E

Chromatic aberration is due to dispersion of light.

The refractive index of glass depends on wavelength: it is larger for violet than for red. A lens therefore brings different colours to slightly different foci, so a white-light image shows coloured fringes. That variation of refractive index with colour is dispersion.

Refraction alone, with one refractive index, would focus all colours at the same point, so 'refraction' is not the specific cause. Scattering, interference and reflection play no part in forming the lens image.

Answer: B

The angle of refraction is 33°.

Reflected light is completely polarized when it strikes at Brewster's angle, given by \(\tan i_{B} = n\). Since \(n = \tan 57^{\circ}\), the angle of incidence is 57°.

At Brewster's angle the reflected and refracted rays are perpendicular, so \(i_{B} + r = 90^{\circ}\) and

\[ r = 90^{\circ} - 57^{\circ} = 33^{\circ} \]

90° is the angle between the two rays, not the angle of refraction, and 37° or 45° do not satisfy the Brewster condition.

Answer: A

Light energy is redistributed in diffraction and interference.

In both phenomena waves superpose. Energy is not created or destroyed: it is taken from the dark regions (destructive superposition) and added to the bright ones (constructive superposition), so the average intensity is unchanged while its distribution in space changes into fringes.

Reflection and refraction only change the direction of light, and polarization restricts the direction of vibration; none of them produces a pattern of bright and dark regions by superposition.

Answer: B

The most deviated colour is violet.

The refractive index of glass is highest for the shortest visible wavelengths, so violet light bends the most on passing through a prism and red the least. That is why violet is at the bottom and red at the top of the spectrum from a prism with its apex upwards.

Blue, green and yellow lie between violet and red and are deviated less than violet.

Answer: C

Diffraction is most significant when the slit width is comparable to the wavelength.

The angular spread of the central maximum is about \(\frac{\lambda}{a}\). When the slit width a is close to \(\lambda\), the light spreads into a broad, clearly visible pattern. When the slit is much wider than \(\lambda\), the spreading is negligible and light travels nearly in straight lines.

When the slit is much narrower than \(\lambda\), so little light passes that no useful pattern forms. The screen distance and independence from wavelength have nothing to do with how strongly light diffracts.

Answer: C

At minimum deviation the angle of incidence equals the angle of emergence.

The lowest point of the deviation versus incidence curve is the angle of minimum deviation. There the ray passes symmetrically through the prism: i = e, and the angles of refraction at the two faces are equal, \(r_{1} = r_{2} = \frac{A}{2}\).

So it is the two angles of refraction that each equal half the angle of prism, not the angle of incidence; the incidence angle is larger than either refraction angle because the ray bends towards the normal on entering glass. The angle of incidence is not equal to the angle of prism in general.

Answer: A

The refractive index is \(2\cos\frac{A}{2}\).

The prism formula at minimum deviation is \(\mu = \frac{\sin\frac{A + \delta_{m}}{2}}{\sin\frac{A}{2}}\). With \(\delta_{m} = A\):

\[ \begin{aligned} \mu &= \frac{\sin A}{\sin\frac{A}{2}} \\ &= \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}} \\ &= 2\cos\frac{A}{2} \end{aligned} \]

\(\cos\frac{A}{2}\) drops the factor 2 from the double-angle formula. \(2\cos A\) and \(\cos A\) use the full angle, and \(\sin\frac{A}{2}\) is the denominator, not the result.

Answer: A

The fringe width triples if the wavelength is tripled.

In Young's experiment the fringe width is:

\[ \begin{aligned} \beta &= \frac{\lambda D}{d} \end{aligned} \]

With the same set-up, \(D\) and \(d\) are fixed, so \(\beta \propto \lambda\), and tripling \(\beta\) needs tripling \(\lambda\), to 1800 nm. (The given 600 nm, 0.5 mm and 2 m set the original \(\beta = 2.4\) mm but are not needed for the ratio.)

Making \(\lambda\) one-third would make the fringes one-third as wide, the inverse of what is asked. Doubling or halving changes \(\beta\) by the same factor of 2.

Answer: E

Astigmatism is corrected with a cylindrical lens.

In astigmatism the cornea is not spherical: it has different curvature in different planes, so lines in one direction are focused while those at right angles are not. A cylindrical lens has curvature in one plane only and corrects the faulty plane.

Convex and concave (spherical) lenses correct hypermetropia and myopia, which affect all planes equally, so they cannot fix astigmatism alone.

Answer: A

The focal length is R.

By the lens maker's formula, for a biconvex lens with \(R_{1} = R\) and \(R_{2} = -R\):

\[ \begin{aligned} \frac{1}{f} &= (\mu - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \\ &= \left(\tfrac{3}{2} - 1\right)\frac{2}{R} = \frac{1}{R} \end{aligned} \]

So \(f = R\). The option 2R uses only one surface, as for a plano-convex lens of the same glass.