The polarising angle for a piece of glass for green light is `60^(@)`. Find the angle of minimum deviation for green light for its passage through `60^(@)` prism, made of the same glass. Given 1sin `60^(@) = 0.866, tan 60^(@) = 1.732`.
Text Solution
Verified by Experts
Here, `i_(p) = 60^(@), delta_(m) = ? A = 60^(@)` As `mu = tan i_(p) = tan 60^(@) = sqrt(3)` From prism formula, `mu = (sin(A + delta_(m))//2)/(sin A//2)` `sqrt(3) = (sin(60^(@) + delta_(m))//2)/(sin 30^(@))` `(sin(60^(@) + delta_(m)))/(2) = sqrt(3) xx (1)/(2) = sin 60^(@)` `(60^(@) + delta_(m))/(2) = 60^(@) delta_(m) = 2 xx 6^(@) - 60^(@) = 60^(@)`
Topper's Solved these Questions
OPTICS
PRADEEP|Exercise Very short answer question|5 Videos
OPTICS
PRADEEP|Exercise very short answer questions|1 Videos
If the polarizing angle of a piece of glass for green light is 54.74^(@) , then the angle of minimum deviation for an equilateral prism made of same glass is : ["Given" : tan 54.74^(@)=1.414]
The refracting angle of a glass prism is 60^(@) . The value of minium deviation for light passing through the prism is 40^(@) . Calculate the value of refractive index of the material of the prism. Given sin 50^(@) = 0.766 .
The refraction index of a prism of angle 60^(@) is 1.62 for sodium light. What is the angle of minimum deviation.
The angle of a prism is 60^(@) and its refractive index is sqrt2 . The angle of minimum deviation suffered by a ray of light in passing through it is
The angle of a prism is 60^(@) and its refractive index is sqrt(2) . The angle of minumum deviation sufferred by a ray of light in passing through it is
If the angle of prism is 60^(@) and the angle of minimum deviation is 40^(@) , the angle of refraction will be
If the polarising angle for red light is 60^(@) , for a certain medium, then the polarising angle for the blue light for the same medium will be
