Question #354d2

2 Answers
Jan 14, 2018

41.81 degrees

Explanation:

enter image source here Critical angle of light is that one,if at that angle light falls at an interface,while going from a solid to lighter media,it will not enter the media next to the interface,rather go along the interface,

So,using Snell's law we can say,
#u1 sin theta# = #u2 sin alpha#
Given, #u2# = #(3/2)#
And #sin theta# = 1
So, #sin alpha# =# 2/3# or #alpha# = 41.81 degrees

Jan 14, 2018

the critical angle is #i_c=42.067#

Explanation:

critical angle #i_c# is given as
#sin(i_c)=1/n_21# where #n_21=n_2/n_1#
#n # is the refractive index of that particular media
#here n_2=3/2 and n_1=1# substituting in the formula we get
#sin(i_c)=(1/(3/2))=(2/3)rArri_c=sin^-1(2/3)=sin^-1(0.67)#
#rArri_c=42.067#