Question

You are scuba diving underwater (n = 1.33) near an old shipwreck, and come upon a very sparkly stone. Could it be a diamond (n = 2.42)? You have a laser handy, and shine it from (he water into the diamond at an angle 25degree relative to the normal. a. What angle should the beam make when it retracts into the diamond? b. To get a good idea of your depth, you shine the laser at the surface of the water. You don't want to accidentally shine it in the eyes of anyone on the boat above, so you have to make sure all the light is reflected back into the water. What is the minimum angle for (his to occur? (Assume n = 1 for air.)

Answer 1

Part a) according to the snell's law we know that if light is passing from one medium of refractive μ₁ to another medium of refractive index μ₂ then according to the law

μ₁ sin i= μ₂ sin r

it is given in the question that light is passing from water to diamond

therefore our μ₁ =μwater= 1.33

and μ₂ =μ_diamond= 2.42

and angle of incidence is given as i=25^o

μ₁ sin i= μ₂ sin r

1,33 sin25^o= 2.42 sin r

sin r=1.33 sin25^o/2.42

sin r=(1.33 *0.4226182)/2.42

sinr=0.232265

r= sin^-10.232265

r=13.430 ^o

This is the angle at which ray refract inside diamond

Part B) in order to make the light reflect back to water itself we need to have total internal reflection

therfore angle of refratio would be 90⁰ which is nothing but the critical angle

according to snells law

μ₁ sin i= μ₂ sin r

μ₁ =μwater= 1.33

μ2 =μair= 1

sin r=sin 90⁰=1

therfore 1.33 sini =1 *1

sin i=1/1.33

sini=0.751879

i= sin^-10.751879

i=48.75⁰

This is the minimum angle at which ray reflect back to water

Part c) speed of light in air=3 * 10⁸ m/s

speed of light in any medium is given by =speed of light in air/refractive index of the medium

speed of laser beam in diamond=3 * 10⁸/2.42

speed of laser beam in diamond=1.23966 *10⁸ m/s

speed of laser beam in water=3 * 10⁸/1.33

speed of laser beam in water=2.2556 *10⁸ m/s

GOOD LUCK!!

^{angle at which light will reflect back}