Question

critical Angle for light moving from medium 1 to medium 2 is theta. the speed of light in medium 1 is v then the speed in medium 2 is

Answer 1

From Snell's Law:
n₁sini=n₂ sinr-------------equation (1)
where n1 =Refractive index of first medium
Refractive index of second medium
i=angle of incidence
r- angle of refraction

For critical angle::
Angle of incidence=i=θ
angle of refraction=r=90°
by substituting in equation 1 we get :
n₁sin θ=n₂sin90
n₁sin θ=n₂x1
sin θ=n₂/n₁--------------2

But as we already know the equation= n=c/V
where n inversely proportional to V
if v and v2 are the speeds of first medium and second medium
then :

n₂/n₁=v/v2---------------3
by equation 2 and 3 we get,
sin θ=v/v2
v2sin θ=V
⇒v2=V/sinθ

Answer 2

we know, from Snell's law
$\bold{\mu_i sin(\theta_i)=\mu_r sin(\theta_r)}$
where μ show refractive index of mediums and θ show angle of incidence and refracted.

for critical angle ,
$\bold{\theta_i=\theta_{cr}}$
$\bold{\theta_r=90}$

$\bold{then, sin{\theta_{cr}}=\frac{\mu_r}{\mu_i}}$

but we know, speed of light is inversely proportional to refractive index of medium,
$e.g., \frac{v_i}{v_r}=\frac{\mu_r}{\mu_i}$
$so \: \: \: \: \bold{sin{\theta_{cr}}=\frac{v_i}{v_r}}$
here, critical angle is theta , vi = v1 = v
then, vr = v2

$\bold{sin{\theta}=\frac{v}{v_2}}$
$\bold{v_2 =\frac{v}{sin{\theta}}}$