Question

Commonly used glass have refractive index of 1.5 the critical angle for such glass is:
(a)49 ° (b)42°
(c)45° (d)40°​

Answer 1

Given :

The refractive angle of glass = 1.5

To Find :

The critical angle ($i_{c}$)

Solution :

Let the ray of light travelling from glass to air.

Let μ be the refractive index of glass.

Using Snell's Law,

             sin$i_{c}$ × μ = sin90° × 1

or,       sin$i_{c}$ × 1.5 = 1

or,              sin$i_{c}$  = $\frac{1}{1.5}$

or,                  $i_{c}$   = $sin^{-1} (\frac{2}{3} )$

∴                    $i_{c}$   =  41.8° or 42°

∴ The critical angle for glass is (b) 42°.

The statement, 'critical angle of glass relative to air is 42°' means that a ray of light from glass being incident on the surface of separation of glass and air at an angle 42°, should go along the surface after refraction i.e., the refracted angle will be 90°.

Answer 2

The critical angle for the glass will be (b) 42⁰.

Given:

Refractive index of glass (μ₂) = 1.5

To find:

The critical angle ($i_{c}$) for this glass medium.

Explanation:

We know,

• The second law of refraction or Snell's law states that, " For a given pair of media, the ratio of sine of angle of incidence to the sine of angle of refraction is a constant ". This constant is nothing but the Refractive Index of the pair of media.

                 Mathematically,  μ $=sin i/ sin r$

• A critical angle is a very distinct angle at which, a ray of light travelling through one medium after refraction gets deviated at an angle of 90° or the angle of refraction in this case is 90°.

Solution:

We have been given that a ray of light travelling from the air medium (μ₁) whose refractive index is 1, towards the glass medium (μ₂) whose refractive index is 1.5.

And, since we have been asked critical angle hence, here the angle of refraction  here will be 90°.

Hence, substituting this information in Snell's law, we get

μ₁/μ₂$= sin i/sin r$  ; and

$\frac{1}{1.5}=\frac{sin(i_{c} )}{sin(90)}$

and

$\frac{1}{1.5}=\frac{sin(i_{c} )}{1}$

$sin(i_{c} ) =\frac{2}{3}$

Therefore  $i_{c}=sin^{-1}(\frac{2}{3} )$

or $i_{c} = 42^{o}$

Final answer:

Hence, the critical angle for the glass medium is (b) 42°.