Question

Suppose a glass cube of edge $1cm$ and $\mu =1.5$ has a spot at the center a. Then find the area of the cube face that must be covered to prevent the spot from being seen is (in c.m.) –
A. $\sqrt{5\pi }$
B. $5\pi$
C. $\dfrac{\pi }{\sqrt{5}}$
D. $\dfrac{\pi }{5}$

Answer 1

In the case of light of a given color and given media we learn from Snell’s law that the ratio of sine of incident angle to the refraction angle is constant .We have to calculate the angle of incident and then find the area of cube.

Complete step-by-step solution:
According to the question, the glass cube has $1cm$edge and $\mu =1.5$at the center.
So, the angle of incident
${{n}_{1}}\sin {{\theta }_{1}}={{n}_{2}}\sin \theta$
$\Rightarrow 1\sin {{90}^{\circ }}=1.5\sin \theta$
$\Rightarrow \sin \theta =\dfrac{1}{1.5}=\dfrac{2}{3}$
$\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{2}{3} \right)$
$\Rightarrow \theta =41.8$
Radius of the circle $\Rightarrow \tan =\dfrac{r}{{}^{a}/{}_{2}}$
$\Rightarrow \tan 41.8=\dfrac{2r}{1}$ [ putting the value of $\theta$ and $a$]
$\Rightarrow 0.8941=2r$
$\Rightarrow r=0.45cm$
Assuming the area is A , calculate covered area $\Rightarrow A=6\left( \pi {{x}^{2}} \right)$
$\Rightarrow A=6\times \left( \pi {{\left( 0.45 \right)}^{2}} \right)$

& \Rightarrow A=3.77c{{m}^{2}} \\\ & \Rightarrow A=\dfrac{\pi }{5}c{{m}^{2}} \\\ \end{aligned}$$ **Therefore, the right answer is option D) $$\dfrac{\pi }{5}$$.** **Note:** Reflection is governed by the equations $$\angle i=\angle r'$$ and reflection by the Snell’s law , $$\dfrac{\sin i}{\sin r}=n$$ , where the incident ray , refracted ray , reflected ray lies in the same plane. The critical angle for incident from denser to rarer medium is that angle for which the angle of refraction is $${{90}^{\circ }}$$.