Reflection and Refraction of Light

1. A parallel beam of light having a width of 10 cm is incident from the air on a water surface ($n =1.33$) under an angle $hat i = 60 deg$. Find the width of the refracted beam that enters the water.

2. Show that is a plane mirror is moved a certain distance $d$ then the position of the image of a same object is moving with twice the distance $d$.

3. A beam of light falls under a certain incidence angle on a transparent sphere made from a material with refraction index $n$. Find this incidence angle if there is maximum deviation of the beam that emerges out from the sphere.


Show My Homework - Refraction

From the geometry of figure above one has $L_2 /cos(hat r)=L_1/cos(hat i)$.

From the Snell’s law we have: $n =sin(hat i)/sin(hat r)$

$hat r = arcsin(sin(hat i)/n) = arcsin(sin(60)/1.33) =40.63 deg$

$L_2 =L_1*cos(40.63)/cos(60)=  15.18 cm$


Show My Homework - Reflection 1

Initially the mirror is in the $K$ position. It is moved with a distance $d$ to $K’$ position.

From the geometry of the figure we have

$|A_1A_1’| =|A_1’A| -|A_1A|=2*(|AK’| -|AK|) =2d$


Show My Homework - Reflection 2

From the geometry of the figure we have:

$hat r=hat i -hat r +hat a/2$

or equivalent $hat a =4hat r-2hat i$

From the refraction law we know that: $hat i =arcsin [n*sin(hat r)]$ so that

$hat a =2[2hat r-arcsin[n*sin(hat r)]$

condition of extrema of $hat a$ is

$frac{d(hat a)}{d r}=2left [ 2-frac{n*cos(hat r))}{sqrt{1-n^2*sin^2(hat r)}} right ] =0$

which gives

$sin(hat r)  =sqrt{(4-n^2)/(3n^2)}$

Therefore the incidence angle for maximum deviation inside the sphere is:

$(hat i)_max =arcsin (sqrt{(4-n^2)/3})$