Question

a) What is a mirror's radius of curvature if cars 17.0 m away appear 0.27 their normal size? Follow the sign conventions. b) A diver shines a flashlight upward from beneath the water (n = 1.33) at a 31.2 ∘ angle to the vertical. At what angle does the light leave the water?

Answer 1

The ratio between the height(size) of the image and the height(size) of the object is equal to the ratio between the image-mirror distance and object-mirror distance.

Here,the image size is 0.27 times the object size.

so, h_i/h_f = d_i/d_o = 0.27

d_o=17m

d_i=0.27d_o

Using the mirror equtaion,we can find the focal length f of the mirror,

1/f=1/do + 1/d_i

Put d_i=0.27*17= 4.59m in the mirror equation

we find f is equal to 3.84m

radius of curvature=2(focal length)=2f= 7.7m

Now coming to part b

From snell's law

sin i/sin r = n₂/n₁ {n₂ is refractive index of water and n₁ is refractive index of air}

sin 31.2/sin r = 1.33/1 (Assuming refractive index of air as 1)

r=43.47^o (from the vertical)