Question

A block of crown glass of refractive index 1.52, is immersed in
water of refractive index 1.33, as in the figure shown below. A
light ray is incident on the top face at an angle of
*θ*_{1} = 42.2° with the normal and exits the block
at point *P*. (Assume that *x* = 3.63 cm.)

A block is immersed in water. A ray starts in the water at the
top of the image and moves down and to the right before it is
incident on the top surface of the block at an angle
*θ*_{1} to the vertical at a distance *x*from
the right side of the block. The ray refracts to have a steeper
slope than the incident ray, and extends down and to the right to
point *P* on the right side of the block, where point
*P* is a vertical distance *y* from the top of the
block. The ray exits the block at a steeper angle than it was
within the block, and at an angle *θ*_{2} to the
horizontal.

(a) Find the vertical distance *y* from the top of the
block to *P*.

cm

(b) Find the angle of refraction *θ*_{2} of the
light ray leaving the block at *P*.

°

I WILL RATE

Answer #1

a)

applying Snell’s law of refraction at the top surface,

n1 Sinθ1 = n2 Sinθ

1.33 x Sin 42.2 = 1.52 x Sinθ

Angle of refraction at the top surface, θ = 36^{0}

Tan θ = x/y

Vertical distance, y = x/tan θ = 3.63/tan 36 = 5.0 cm

The vertical distance, y = 5.0 cm

b)

the angle of incidence on the vertical side at P is I = 90 – θ =
90 – 36 = 54^{0}

applying Snell’s law of refraction at the vertical surface at P,

n2 Sin i = n1 Sinθ2

1.52 x Sin 54 = 1.33 x Sinθ2

Angle of refraction at the vertical surface, θ2 =
67.6^{0}

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