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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 $
A ray of light is incident normally on one of the faces of a solid prism of angle $ 30 ^\circ $ and refractive index $ \sqrt 2 } . The angle of minimum deviation is
$ here , i = 90 ^ \circ , r_1 = 0 , r_1 + r_2 = A , r_2 = 30 ^\circ $ $ Now , n = { sin (i_2 ) \over sin (r_2) } \therefore i_2 = 45 ^\circ $ $ i + e = A + \delta m $ $ \therefore \delta m = 15 ^ \circ $
A concave mirror has a focal langth 30 cm The distance between the two position of the object for whi ch image size is double of the object is
$here : For \, real \,image u = -\nu_1 , \nu = 2 v_1 , f = -30 cm $ $ \therefore { -1 \over -2 u_1} - { 1 \over u_1} = { 1 \over -30 } $ $ \therefore u_1 = 45 cm $ $ for \,virtual\, image\, u = - u_2 , \nu = + 2 \nu_2 , f = -30 cm $ $ \therefore { -1 \over u_2 } + { 1 \over 2 u_2 } = -{ 1 \over 30 } , u_2 = 15 cm $ $ u_1 - u_2 = 30cm $
A concave lens forms the image of an object such that the distance between the object and the image is 10 cm and the magnification produced is 1/4 , the focal length of lens will be ________cm
$ here , m = {1 \over 4 } = { \nu \over u } $ $ \therefore u = 4v $ $ if \nu = - x , u = -4x then $ $ from figure , | 0I | = 4x - x , 3x =10 cm \therefore x = {10 \over 3 } cm $ $ now, u = 4 \nu = 4x ( from fig : \nu = x ) $ $ = + { 40 /3 } and \nu = { u \over 4 } = { 40 \over 3 \times 4 } = { -10 \ove 3 } cm $ $ \therefore from { 1 \over f } = { 1 \over u } - { 1 \over \nu } $ $ \therefore f = - 4.4 cm $
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