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A parallel beam of light is incident on a face of a prism of refracting angle 60^@.Find the refractive index of the prism if the angle of minimum deviation is 40. What is the new angle of minimum deviation if the prism is immersed in water of refractive index 1.33? |
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Answer» Solution : Given, `A = 60^@ , N = ? , D = 40^@ , D^1 = ? , D_w = 1.33` The REFRACTIVE index of the material of a prism is given by CASE (i) `n = (sin ((A+D)/(2)) )/(sin (A/2) ) = (sin ( (60+40)/(2)) )/(sin (60/2))` `n = (sin (50))/(sin (30)) = (0.766)/(0.5)` Case (ii) `(n_g)/(n_w) = (sin ((A+D^1)/(2)) )/(sin(A/2))` `(1.532)/(1.33) = (sin ((60+D^1)/(2)))/(sin (60/2))` `1.151 = (sin ((60+D^1)/(2)))/(0.5)` `1.151 xx 0.5 =sin ((60+D^1)/(2)) ` `0.5755 = sin ((60 + D^1)/(2))` `(60+D^1)/(2) = sin^(-1) (0.5755)` `60+D^1 = 2 xx sin^(-1)(0.5755)` `60+D^1 = 2 xx 35^@ g^1` ` = 70^@ 16^1` `D^1 = 70^@ 16^1- 60^@` `D^1 = 10^@ 16^1` |
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