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.
1.
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: $n =sin(hat i)/sin(hat r)$
$hat r = arcsin(sin(hat i)/n) = arcsin(sin(60)/1.33) =40.63 deg$
$L_2 =L_1*cos(40.63)/cos(60)= 15.18 cm$
2.
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
$|A_1A_1’| =|A_1’A| -|A_1A|=2*(|AK’| -|AK|) =2d$
3.
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:
$(hat i)_max =arcsin (sqrt{(4-n^2)/3})$