Question

3.
For an angle of incidence of 45° at the top surface of a glass slab as shown in figure, what is
the maximum refractive index of glass for total internal reflection? For water u =3/
4
412
(1) 312
(2)
3
3
312
(3) 42
(4)
4​

Answer 1

Answer:

The answer will be √3/2

Explanation:

According to the figure, if we apply Snell's law:

μ == sin i/sin r

Now it is stated that the angle of incidence is 45 degree

Therefore, sin r = sin i/μ  = 1/μ√2

The total internal reflection sin i1 = 1/μ  

Therefore we can also say that , i1 = 90 -r

Therefore sin(90-r) = 1/μ  

              => cos r = 1/μ  

cos r = √1- sin^2r = √1-1/2μ^2 = √2μ^2-1/2μ^2

from here we can get μ = √3/2

Answer 2

Answer:

4√2/3

Explanation:

Given, Uw=4/3

Sini=45*

Sinr=90* ( TIR for critical angle is 90* )

Now, UgSin1=UwSin2

UgSin45*=4/3Sin90*

Ug.1/√2=4/3.1

Ug=4/3.√2/1

Ug=4√2/3

Solved!!!