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When a picture drawn on a paper is seen through a slab of a transparent material of thickness $5cm$, it appears to be raised by $1.5cm$. Critical angle at the boundary of this transparent material and air is
A. ${\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
B. ${\sin ^{ - 1}}\left( {\dfrac{5}{7}} \right)$
C. ${\sin ^{ - 1}}\left( {\dfrac{6}{{11}}} \right)$
D. ${\sin ^{ - 1}}\left( {\dfrac{7}{{10}}} \right)$

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Last updated date: 26th Feb 2024
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Hint Refraction of light changes depth perception. Taking the thickness of the slab to be the real depth of the picture, calculate the refractive index of the material of the slab. Then using the formula for critical angle, obtain the solution.
Formula used
$n = \dfrac{d}{{d'}}$ where $d$ is the real depth, $d'$ is the apparent depth and $n$ is the refractive index of the medium.
$\sin {i_c} = \dfrac{1}{n}$ where ${i_c}$ is the critical angle.

Complete step by step answer
When light travels from one medium to another its direction changes. This optical phenomenon is known as the refraction of light. Now the degree up to which light bends as it changes medium is known as the refractive index. The refractive index $n$ for a given medium is given as
$n = \dfrac{c}{v}$ where $c$is the velocity of light in vacuum and $v$ is the velocity of light in the given media.
Another property of refraction is that when an object placed in a medium is viewed from another medium, its distance appears to be different than the original distance. As given in the question above, when a picture drawn on paper kept below a rectangular slab is observed from above (different medium), its depth appears to have changed. This virtual depth is known as apparent depth and the original thickness of the slab here is the real depth.
Another formula to calculate refractive index is
$n = \dfrac{d}{{d'}}$ where $d$ is the real depth and $d'$ is the apparent depth.
Now we are given, $d = 5cm$ and $d' = \left( {5 - 1.5} \right)cm = 3.5cm$
Therefore, $n = \dfrac{5}{{3.5}} = 1.428$
So, the refractive index of the material of the slab with respect to air is $1.428$.
The critical angle comes into play when light travels from denser to rarer medium. Critical angle is the angle of incidence when the angle of refraction is ${90^ \circ }$, which means that the refracted ray lies along the boundary.
The expression for critical angle is $\sin {i_c} = \dfrac{1}{n}$ where ${i_c}$ is the critical angle.
$\begin{gathered}
  \sin {i_c} = \left( {\dfrac{{3.5}}{5}} \right) \\
   \Rightarrow {i_c} = {\sin ^{ - 1}}\left( {\dfrac{{35}}{{50}}} \right) \\
   \Rightarrow {i_c} = {\sin ^{ - 1}}\left( {\dfrac{7}{{10}}} \right) \\
\end{gathered} $

Therefore, the correct option is D.

Note Critical angle is usually defined as the largest angle of incidence at which refraction can still occur. If the angle of incidence is greater than the critical angle for that medium, then total internal reflection will take place.