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The magnifying power of a telescope is 9.0 when it is focussed for parallel rays, then the distance between its objective and eye–piece is 20 cm The focal lengths of lenses will be
A plano convex lens of f = 20 cm is silvered at plane surface New f will be....... cm
$ use { 1 \over f } = ( n -1 ) \left( { 1 \over R_1 } - { 1 \over R_2 } \right) $ $ \therefore R = 10 cm $ $ for \,rarer \,medium \,to\, denser , -{ n_1 \over u } + { n_2 \over \nu } = { n_2 - n_1 \over R } $ $ ( \therefore u = \infty , v = f ) $ $ \therefore { 0 + 1.5 \over f } = { 1.5 - 1 \over 10 } $ $ \therefore f = 30 cm $
A ray of light from denser medium strikes a rarer medium at angle of incidence i. The reflected and refracted rays make an angle of $90 ^\circ $ with each other The angle of reflection and refration are r and r' respectively. The crictical angle is
Relation between critical angle of water $C_w$ and that of the glass $C_g$ is ..........$( given , n_w = 4/3 , n_g = 1.5 )$
$ here , Cw = sin ^ {-1} \left( { 1 \over n_w} \right) = sin ^ {-1} \left( { 3 \over 4 } \right) = 48. 6 ^ \circ $ $ Cg = sin ^ {-1} \left( { 1 \over ng } \right) = 42 ^\circ $ $ \therefore Cw \gt Cg $
The radius of curvature of convex surface of a thin plano–convex lens is 15 cm and refractve index of its material is 1.6 The power of the lens will be
$ use { 1 \over f } = (n-1) \left( { 1 \over R_1 } - { 1 \over R_2 } \right) $ $ \therefore f = 0.25 m , P = 4D $
A ray of light passes through a prism having refractive index $( n = \sqrt 2) $ , Suffers minimum deviation If angle of incident is double the angle of refration within prism then angle of prism is
$ use \mu = { sin i \over sin p } = { sin 2r \over sin r } = { 2 sin r . cos r \over sin r } = 2 cos r $ $ \therefore cos = { \sqrt 2 \over 2 } = { 1 \over \sqrt 2 } $ $ r = 45 ^ \circ , \therefore A = 90 ^ \circ $
An air bubble inside glass slab (n =1.5) appear from one side at 6 cm and from other side at 4 cm. Then the thickness of glass slab is_ _cm
$ use n = { Real depth \over Apparent \, depth } = { x \over y} $ $ \therefore x = ny = 15 cm $ $ ( \therefore Apparent \,depth = 6 + 4 = 10 ) $
The magnifying power of objective of a compound microscope is 5.0 If the maginfying power of microscope is 30, then magnifying power of eye–piece will be .
The magnifying power of a compound microscope (M) is given by the product of the magnifying powers of the objective (M_o) and the eyepiece (M_e). So, M = M_o * M_e. Given M = 30 and M_o = 5, we can find M_e by solving 30 = 5 * M_e, which gives M_e = 6.
Light of certain colour contain 2000 waves in the length of 1 mm in air. What will be the wavelength of this light in medium of refractive index 1.25 ?
$ use = { distant \over wave no. } = { 5000 A ^ \circ } $ $ now \,\lambda' = { \lambda \over n } = 4000 A ^ \circ $
A convex lens of glass (n =1.5) has focal lergth 0.2 m The lens is immersed in water of refractive index 1.33. The change in the power of convex lens is
$ here , wn _g = an_g / an_g = 1.128 $ $ Now , { 1 \over fa } = ( an_g -1 ) \left( { 1 \over R_1 } - { 1 \over R_2 } \right) $ $ \therefore {1 \over R_1} - { 1 \over R_2 } = 10 $ $ and { 1 \over fw} = ( wn_g -1 ) \left ( { 1 \over R_1 } - { 1 \over R_2 } \right) = (1.128 -1 ) \times 10 = 1.28 $ $ then \therefore Pa = { 1 \over fa} = 5D \, and Pw = 01.28 $ $ \therefore Pa - Pw = 3.72 D $
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