Question

A ray of light strikes a flat glass block at an incidence angle of θ1 = 34.2°. The glass is 2.00 cm thick and has an index of refraction that equals ng = 1.76.

PICTURE: A light ray incident on a glass block of thickness 2.00 cm is shown. The ray travels down and to the right and is incident to the top of the block at an angle θ1 to the normal of the surface. The ray inside the block moves down and to the right but at a steeper slope than the incident ray, making an angle of θ2 with the vertical. It is incident on the bottom surface of the block, making an angle of θ3 with the vertical, and exits moving down and to the right, at a less steep slope, making an angle of θ4 with the vertical. A dashed line extends from the original path of the ray down in the block and is shown to be a distance d from the ray that exits the glass block.

a.) What is the angle of refraction, θ_{2}, that
describes the light ray after it enters the glass from above?
(Enter your answer in degrees to at least 2 decimal places.)

b.) With what angle of incidence, θ_{3}, does the ray
*approach* the interface at the bottom of the glass? (Enter
your answer in degrees to at least 2 decimal places.)

c.) With what angle of refraction, θ_{4}, does the ray
*emerge from* the bottom of the glass? (Enter your answer in
degrees to at least 1 decimal place.)

d.) The distance *d* separates the twice-bent ray from
the path it would have taken without the glass in the way. What is
this distance (in cm)?

e.) At what speed (in m/s) does the light travel within the glass?

f.) How many nanoseconds does the light take to pass through the glass along the angled path shown here?

g.) Is the travel time through the block affected by the angle of incidence (and if so, how)?

a. Yes, a slightly larger angle will increase the travel time.

b. No, the time taken for the light to traverse the block is independent of incidence angle.

c. Yes, a slightly larger angle will decrease the travel time.

Answer #1

By applying snell's law at surface 1

Sin**θ**1/Sinθ2 = n2 /n1

Sin34.2/Sinθ2 = 1.76/1

**θ2** = 18.6o

At surface 2

**θ2 = θ3**

Sinθ3/Sinθ4 = nair/nglass

Sin18.6/Sinθ4 = 1/1.76

**θ4** = 34.2o

This is also we know that these both rays are parallel. So angle must be same.

Shift (d) = LSin(**θ1-θ2)**

**LCosθ2 = thickness (t)**

L = 2/Cos18.6 = 2.11 cm

d = 2.11 × Sin (34.2 - 18.6)

d = 0.57 cm

e)

n = c/v

v = 3×108/1.76 = 1.7×108 m/s

f.

L = vt

t = 2.11×10-2/1.7×108 = 0.124 nsec

g.

As angle is more distance travelled by light is more so time taken is more

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