Question

A horizontal parallel-sided plate of glass having a refractive index of 1.56 is in contact with the surface tnk. A ray coming from above in air makes an angle of incidence of 32.5. with the normal to the top surface of the glass. Take refractive index of air 1 a) What is the angle of refraction on glass surface? (5 pts) of wa er i b) What is the critical angle of gla ss-water surface as explained above?

Answer 1

Answer:

(I). The angle of refraction on glass surface is 20.12°

(II). The critical angle of glass-water surface is  61.0°.

Explanation:

Given that,

Refractive index of glass = 1.56

Angle = 32.5

(I). We need to calculate the angle of refraction on glass surface

Using Snell's law

$n_{a}\sin i=n_{g}\sin r$

Put the value into the formula

$1\times\sin32.5=1.56\sin\theta$

$\sin r=\dfrac{1\times\sin32.5}{1.56}$

$r=\sin^{-1}0.344$

$r=20.12$

(II). We need to calculate the critical angle of glass-water surface

Using formula of critical angle

$\sin\theta_{c}=\dfrac{n_{w}}{n_{a}}$

$\theta_{c}=\sin^{-1}\dfrac{1.33}{1.52}$

$\theta_{c}=61.0^{\circ}$

Hence, (I). The angle of refraction on glass surface is 20.12°

(II). The critical angle of glass-water surface is  61.0°.