Optics Questions and Answers

Answer: A

Its frequency remains unchanged.

Frequency is fixed by the source: the waves arriving at the boundary must match the oscillations they set up on the other side, so the number of waves per second cannot change. The speed changes to \(v = \frac{c}{n}\), and since \(v = f\lambda\), the wavelength changes in proportion.

So options saying the frequency changes, the wavelength stays the same, or the velocity stays constant are all wrong.

Answer: B

Brewster's angle lies between \(45^{\circ}\) and \(90^{\circ}\).

By Brewster's law \(\tan i_{B} = n\), where \(n\) is the refractive index of the second medium relative to the first. For light going from a rarer to a denser medium (the usual case, such as air to glass), \(n \gt 1\), so \(\tan i_{B} \gt 1\) and \(i_{B} \gt 45^{\circ}\). Since \(\tan\) is finite, \(i_{B} \lt 90^{\circ}\).

For air to glass with \(n = 1.5\), \(i_{B} \approx 56^{\circ}\), inside this range. The ranges starting at \(0^{\circ}\) or ending at \(45^{\circ}\) or below would need \(n \lt 1\).

Answer: C

The wavelength required is \(2\lambda\).

Fringe width is \(\beta = \frac{\lambda D}{d}\). With the same slit separation \(d\) and screen distance \(D\), \(\beta\) is directly proportional to \(\lambda\), so doubling \(\beta\) needs double the wavelength.

\(\frac{\lambda}{2}\) would halve the fringe width, and the smaller fractions would shrink it further.

Answer: B

The power of each piece is \(\frac{P}{2}\).

Cutting a biconvex lens perpendicular to its principal axis splits it into two plano-convex lenses. For an equiconvex lens of radius R, \(\frac{1}{f} = (\mu - 1)\frac{2}{R}\); for each plano-convex half, \(\frac{1}{f'} = (\mu - 1)\frac{1}{R}\). So \(f' = 2f\) and the power halves.

A cut along the axis (parallel to it) would leave the focal length, and so the power P, unchanged; that is the trap in option A. 2P and 4P go the wrong way.

Answer: B

The fringe width becomes 1.0 mm.

Fringe width \(\beta = \frac{\lambda D}{d}\). In a medium of refractive index \(\mu\) the wavelength becomes \(\lambda/\mu\), so

\[ \begin{aligned} \beta' &= \frac{\beta}{\mu} = \frac{1.4}{1.4} \\ &= 1.0\ \text{mm} \end{aligned} \]

1.4 mm ignores the change of wavelength, 0.7 mm divides by 2 instead of 1.4, and 2.8 mm multiplies instead of dividing.

Answer: A

Monochromatic light does not produce dispersion.

Dispersion is the splitting of light into its component colours, which happens because the refractive index of glass is different for different wavelengths, so each colour is bent by a different amount. Monochromatic light has a single wavelength, so there is nothing to separate: it emerges from the prism as one coloured beam.

The other effects do not depend on there being several wavelengths. Monochromatic light is still refracted at each face, and therefore deviated from its path. Some of it is reflected at each surface, and it can be totally internally reflected inside the prism if it meets a face at more than the critical angle.

Answer: B

The magnification is 6.

A simple microscope (a magnifying glass) gives its maximum magnification when the image is formed at the least distance of distinct vision, D = 25 cm:

\[ m = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 6 \]

If instead the image is formed at infinity, for relaxed viewing, the magnification is D/f = 25/5 = 5. The key takes the standard case of the image at the near point, since the question gives the least distance of distinct vision. 4, 10 and 12 do not come from either formula.

Answer: A

The keyed distance is 8 cm.

The focal length is \(f = \frac{R}{2} = 12\) cm. A magnification of \(+3\) means an erect, virtual, enlarged image, formed when the object is inside the focus. With \(m = -\frac{v}{u} = 3\), \(v = -3u\), and the mirror formula with the sign convention gives

\[ \begin{aligned} \frac{1}{v} + \frac{1}{u} &= \frac{1}{f} \\ -\frac{1}{3u} + \frac{1}{u} = \frac{2}{3u} &= -\frac{1}{12} \;\Rightarrow\; u = -8\ \text{cm} \end{aligned} \]

The object is 8 cm in front of the mirror. If the magnification were \(-3\) (a real, inverted image), the distance would be 16 cm instead; the key takes the stated magnification 3 as positive.

Answer: D

The ratio is the refractive index of the medium.

The absolute refractive index is defined as \(n = \frac{c}{v}\), the speed of light in vacuum divided by its speed in the medium. Since light is slowest in dense media, \(n\) is greater than 1 for them.

Luminous intensity and luminous flux are photometric measures of light output, magnification is a property of images, and 'transparency index' is not a standard term.

Answer: A

The angular separation becomes \(2\omega\).

The angular fringe separation in Young's experiment is

\[ \omega = \frac{\lambda}{d} \]

which is inversely proportional to the slit separation \(d\). Halving \(d\) doubles it. \(\frac{\omega}{2}\) would follow from doubling the separation, and the factors \(\sqrt{2}\) do not occur.

The linear fringe width \(\beta = \frac{\lambda D}{d}\) on the screen doubles too, since it is the angular separation multiplied by the screen distance.