A ray of light passing from a glass of water is incident on the glass water interface at ${{65}^{o}}$. If the critical angle for the pair of media is ${{63}^{o}}$:
A) The ray will emerge into the water with a deviation of ${{2}^{o}}$ from the normal.
B) The ray will be refracted into the water with a deviation of ${{2}^{o}}$
C) The ray will be totally internally reflected back into the glass with a deviation of ${{50}^{o}}$
D) The ray will be totally internally reflected back into the glass with a deviation of ${{2}^{o}}$
Answer
576.6k+ views
Hint: It is given in the question that the angle of incident is ${{65}^{o}}$ and the critical angle of the medium is ${{63}^{o}}$ hence we apply the simple principle that is when angle of incidence is greater than critical angle a ray will get totally reflected back.
Formula used:
$\delta +i+r={{180}^{o}}$
Complete step by step solution:
$\to $ Now it is given that in the question that the angle of incident is i = ${{65}^{o}}$
$\to $ And the critical angle of medium pair is c = ${{63}^{o}}$
$\to $ We know that when the angle of incident or incident angle (i) is greater than critical angle the ray will be reflected from glass as shown in the figure hence.
i > c
$\to $ So we can say that the incident angle will be equal to the reflection angle.
$i=r={{65}^{o}}....\left( 1 \right)$
$\to $Now in order to find the deviation $\left( \delta \right)$ we will use the below formula.
$\delta +i+r={{180}^{o}}......\left( 2 \right)$
$\delta $ = angle of deviation
i = angle of incident
r = angle of reflection
$\to $ Now substitute the value of equation (1) in equation (2) we will get
$\begin{align}
& \Rightarrow \delta +{{65}^{o}}+{{65}^{o}}={{180}^{o}} \\
& \Rightarrow \delta ={{180}^{o}}-{{130}^{o}} \\
& \therefore \delta ={{50}^{o}} \\
\end{align}$
$\to $ So the angle of the deviation is ${{50}^{o}}$ hence our correct answer will be option (c) the ray will be totally internally reflected back into glass with the deviation of ${{50}^{o}}$.
Note:
When ray will be reflected totally we can use below formula to find the angle of the deviation.
$\delta =180-2i$
$\delta $ = angle of deviation
i = angle of the incident.
Formula used:
$\delta +i+r={{180}^{o}}$
Complete step by step solution:
$\to $ Now it is given that in the question that the angle of incident is i = ${{65}^{o}}$
$\to $ And the critical angle of medium pair is c = ${{63}^{o}}$
$\to $ We know that when the angle of incident or incident angle (i) is greater than critical angle the ray will be reflected from glass as shown in the figure hence.
i > c
$\to $ So we can say that the incident angle will be equal to the reflection angle.
$i=r={{65}^{o}}....\left( 1 \right)$
$\to $Now in order to find the deviation $\left( \delta \right)$ we will use the below formula.
$\delta +i+r={{180}^{o}}......\left( 2 \right)$
$\delta $ = angle of deviation
i = angle of incident
r = angle of reflection
$\to $ Now substitute the value of equation (1) in equation (2) we will get
$\begin{align}
& \Rightarrow \delta +{{65}^{o}}+{{65}^{o}}={{180}^{o}} \\
& \Rightarrow \delta ={{180}^{o}}-{{130}^{o}} \\
& \therefore \delta ={{50}^{o}} \\
\end{align}$
$\to $ So the angle of the deviation is ${{50}^{o}}$ hence our correct answer will be option (c) the ray will be totally internally reflected back into glass with the deviation of ${{50}^{o}}$.
Note:
When ray will be reflected totally we can use below formula to find the angle of the deviation.
$\delta =180-2i$
$\delta $ = angle of deviation
i = angle of the incident.
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