Light travels through a glass plate of thickness t and refractive index m. If c is the speed of light in vacuum, the time taken by light to travel this thickness of glass is

μtc
^tc⁄_μ
^t⁄_μc
^μt⁄_c

Solution

If f_V and f_R are the focal lengths of a convex lens for violet and red light respectively and F_V and F_R are the focal lengths of concave lens for violet and red light respectively, then we have

f_V< f_R and F_V> F_R
f_V < f_Rand F_V < F_R
f_V > f_R and F_V > F_R
f_V > f_R and F_V < F_R

Solution

A ray of light is travelling from glass to air. (Refractive index of glass = 1.5). The angle of incidence is 50°.The deviation of the ray is

0°
80°
\(50^{\circ}-sin^{-1}\left [ \frac{sin\, 50^{\circ}}{1.5} \right ]\)
\(sin^{-1}\left [ \frac{sin\, 50^{\circ}}{1.5} \right ]-50^{\circ}\)

Solution

^aµ_g= 1.5

∴ 1.5 = cosecC or C = 42°.Critical angle for glass= 42°. Hence a ray of light incident at 50° in glass medium undergoes total internal reflection. δ denotes the deviation of the ray.

∴ δ = 180° – (50° + 50°) or δ = 80°.

The position of final image formed by the given lens combination from the third lens will be at a distance of(f₁= + 10 cm, f₂= – 10 cm and f₃= + 30 cm).

15 cm
infinity
45 cm
30 cm

Solution

Air has refractive index 1.0003. The thickness of air column,which will have one more wavelength of yellow light (6000Å) than in the same thickness of vacuum is

2 mm
2 cm
2 m
2 km

A ray of light traveling in water is incident on its surface open to air. The angle of incidence is q, which is less than the critical angle. Then there will be

only a reflected ray and no refracted ray
only a refracted ray and no reflected ray
a reflected ray and a refracted ray and the angle between them would be less than 180°–2q
are flected ray and a refracted ray and the angle between them would be greater than 180°–2q

Solution

The ray is partly reflected and partly refracted.

∠MOB = 180 – 2θ

But the angle between refracted and reflected ray is

∠POB. Clearly ∠POB is less than ∠MOB.

The graph shows the variation of magnification m produced by a convex lens with the image distance v. The focal length of the lens is

^b⁄_c
^c⁄_b
b
^ab⁄_c

Solution

A thin lens has focal length f, and its aperture has diameter D. It forms an image of intensity I. If the central part of the aperture, of diameter 2D, is blocked by an opaque paper,,the focal length of the lens and the intensity of image will become

^f⁄₂,ⁱ⁄₂
f,ⁱ⁄₄
^3f⁄₄,ⁱ⁄₂
f,³ⁱ⁄₄

In the displacement method, a concave lens is placed in between an object and a screen. If the magnification in the two positions are m₂and m₂(m₁ > m₂), and the distance between the two positions of the lens is x, the focal length of the lens is

\(\frac{x}{m_{1}+m_{2}}\)
\(\frac{x}{m_{1}-m_{2}}\)
\(\frac{x}{(m_{1}+m_{2})^{2}}\)
\(\frac{x}{(m_{1}-m_{2})^{2}}\)

The layered lens as shown is made of two types of transparent materials-one indicated by horizontal lines and the other by vertical lines. The number of images formed of an object will be

Solution

Due to difference in refractive indices images obtained will be two. Two mediums will form images at two different points due to difference in focal lengths.

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