Question

the critical angle for light going for a medium in which wavelength is 4000 angstrom to medium in which its wavelength is 6000 angstrom is​

Answer 1

Answer:

The answer is C=sin^-1(2/3).

Explanation:

we know that

(v1/v2)=(w1/w2)=(u2/u1)

v=velocity in medium

w=wavelength

u =refractive index

u2/u1 = 6000/4000 = 3/2

u1/u2 = 2/3    

u2sin C=u1sin90    Cis critical angle

sinC=u1/u2

 C=sin^-1(2/3)  

Answer 2

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