Question

6. A ray of light travels from a transparent material to air. The beam is refracted at an angle of 25º to the normal. The beam travels at a speed of 2.20 x 108 m/s in the material. Determine the angle of incidence.

7. A ray of light travels from a transparent material to air. The beam is refracted at an angle of 25º to the normal. The beam travels at a speed of 2.20 x 108 m/s in the material. Determine the total internal reflection angle.

8. A 3.0 cm high object is placed in front of a biconvex lens with a focal length of 20.0 cm. The image is formed 35.0 cm from the vertex. Determine the position of the object in centimeters.

10. A 3.0 cm high object is placed in front of a biconvex lens with a focal length of 20.0 cm. The image is formed 35.0 cm from the vertex. Determine the height of the image in centimeters.

Answer 1

6.

As speed is given in the material, so refractive index of material (n) = c/v = 3×10^8/2.2×10^8

= 1.36

n1 = 1, n2 = 1.36, r = 25

So by snell's law

Sini/Sinr = n2/n1

Sini/Sin25 = 1.36/1

i = 35.2 degree

7.

As refractive index (n) = 1.36

Then by snell's law

Sini/Sinr = n2/n1

Sinθc/Sin90 = 1/1.36

θc = 47.17 degree

8.

Focal length (f) = 20 cm

Image distance (V) = 35 cm

By lens formula

1/v - 1/u = 1/f

1/35 - 1/u = 1/20

u = - 46.67 cm

This is the object distance from mirror.

9.

By magnification = hi/ho = v/u

hi/(3) = 35/(-46.67)

hi = 2.25 cm

This is the height of the image.

If you any doubts regarding this feel free to ask.