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$T h e r e i s a n e q u i - c o n v e x g l a s s l e n s w i t h r a d i u s o f e a c h f a c e a s R a n d {}_{a}μ_{g} = \frac{3}{2} a n d {}_{a}μ_{w} = \frac{4}{3} . I f t h e r e i s w a t e r i n o b j e c t s p a c e a n d a i r i n i m a g e s p a c e, t h e n t h e f o c a l l e n g t h i s :$

$\frac{3}{2} R$

$R^{2}$

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Explanation

For an equi-convex lens with radii R, the focal length is given by f = R/(Î¼g - Î¼w), where Î¼g and Î¼w are the refractive indices of glass and water respectively. Substituting the given values, we get f = (3/2)R.

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A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical. When sunlight falls normally on the mirror, it is focused at distance of 32 cm from the mirror. If the tank is filled with water $(μ = \frac{4}{3})$ upto a height of 20 cm, then the sunlight will now get focussed at

16 cm above water level

9 cm above water level

24 cm below water level

9 cm below water level

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Explanation

When a concave mirror is immersed in a denser medium like water, the focal length increases due to the decrease in the relative refractive index. Using the lens maker's formula, the new focal length can be calculated, which is 9 cm above the water level.

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If the focal length of the objective lens and the eye lens is 4 mm and 25 mm respectively in a compound microscope. The length of the tube is 16 cm. Find its magnifying power for relaxed eye position

32.75

327.5

0.3275

None of the above

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Explanation

(b) $m_{\infty} = \frac{(L - f_{0} - f_{e}) D}{f_{0} f_{e}} = \frac{(16 - 0.4 - 2.5) \times 25}{0.4 \times 2.5} = 327.5$

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An air bubble in a sphere having 4 cm diameter that appears 1 cm from the surface nearest to the eye when looked along diameter. If ${}_{a}μ_{g}$ = 1.5, the distance of bubble from the refracting surface is

1.2 cm

3.2 cm

2.8 cm

1.6 cm

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A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance d in cm will be

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Explanation

For a parallel beam of light to pass through a combination of lenses as a parallel beam, the combination should act as a lens of infinite focal length. This condition is satisfied when the distance between the lenses is equal to the sum of their focal lengths.

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For a convex lens, the distance of the object is taken on X-axis and the distance of the image is taken on Y-axis, the nature of the graph so obtained is :

Straight line

Circle

Parabola

Hyperbola

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Explanation

(d) The graph between the object distance and image distance is a hyperbola.

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The diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If the speed of light in the material of the lens is $2 \times 10^{8} \frac{m}{s e c}$ , the focal length of the lens is

15 cm

20 cm

30 cm

10 cm

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The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image is at a distance of 26 cm from the convex lens, calculate the new size of the image.

1.25 cm

2.5 cm

1.05 cm

2 cm

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An achromatic prism is made by crown glass prism $(A_{c} = 19 °)$ and flint glass prism $(A_{F} = 6 °)$ . If ${}_{C}μ_{v} = 1.5$ and ${}_{F}μ_{v} = 1.66$ , then resultant deviation for red coloured ray will be

1.04°

5°

0.96°

13.5°

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Explanation

(d) For achromatic combination $ω_{C} = - ω_{F}$

${[(μ_{v} - μ_{r}) A]}_{C} = - {[(μ_{v} - μ_{r}) A]}_{F} \Rightarrow {[μ_{r} A]}_{C} \pm {[μ_{r} A]}_{F} = {[μ_{v} A]}_{C} + {[μ_{v} A]}_{F} = 1.5 \times 19 + 6 \times 1.66 = 38.5$
Resultant $ẟ = {[(μ_{r} - 1) A l]}_{C} + {[(μ_{r} - 1) A]}_{F}$

= ${[μ_{r} A]}_{C} + {[μ_{r} A]}_{F} - (A_{C} + A_{F}) = 38.5 - (19 + 6) = 13.5 °$

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A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $μ$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is

$\sin^{- 1} (\frac{1}{μ})$

$\tan^{- 1} (\frac{1}{μ})$

$\sin^{- 1} (\frac{μ - 1}{μ})$

$\cos^{- 1} (\frac{1}{μ})$

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