Solveeit Logo

Question

Physics Question on Ray optics and optical instruments

A ray of light is incident at an angle ii on a glass slab of refractive index μ\mu. The angle between reflected and refracted light is 9090^{\circ}. Then, the relationship between ii and μ\mu is

A

i=tan1(1μ)i=\tan ^{-1}\left(\frac{1}{-\mu}\right)

B

tani=μ\tan \, i = \mu

C

sini=μ\sin \, i = \mu

D

cosi=μ\cos \, i = \mu

Answer

tani=μ\tan \, i = \mu

Explanation

Solution

As situation can be diagrammatically as below

From law of reflection
i=θi=\theta
Now, θ+I+90=180\theta+I+90^{\circ}=180^{\circ}
i+r+90=180\Rightarrow i+r+90^{\circ}=180^{\circ}
r=90ir=90^{\circ}-i
Also, from Snell's law
sinisinr=μ\frac{\sin i}{\sin r}=\mu
sinisin(90i)=sinicosi=μ\Rightarrow \frac{\sin i}{\sin \left(90^{\circ}-i\right)}=\frac{\sin i}{\cos i}=\mu
tani=μ\Rightarrow \tan i=\mu