Answer
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Hint In this solution, we will use the formula for calculating the refractive index of a prism. Using the given values of refractive index and angle of minimum deviation in the question we will get the angle of the prism.
In this solution we will be using the following formula,
$\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$ where $\mu $ is the refractive index, $A$ is the angle of the prism and ${\delta _m}$ is the angle of minimum deviation.
Complete step by step answer
The minimum deviation ${\delta _m}$ in a prism occurs when the entering angle and the exiting angle are the same. In this situation, the ray of light when inside the prism is parallel to the base of the prism. The angle of refraction inside the prism on either side of the prism is also the same in this situation.
If we are given that angle of minimum deviation and the angle of the prism, we can find the refractive index of the prism $\mu $ as:
$\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
Since we have been given in the question as${\delta _m} = A$, we can substitute it in the above equation and write
$\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + A}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
$\Rightarrow \mu = \dfrac{{sin\left( A \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
Since $\mu = 1.5$, we have
$\Rightarrow 1.5 = \dfrac{{sin\left( A \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
Now, we know that $\sin (A) = 2\sin (A/2)\cos (A/2)$ so we can write
$\Rightarrow 1.5 = \dfrac{{2\sin (A/2)\cos (A/2)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
$\Rightarrow 1.5 = 2\cos (A/2)$
On dividing both sides by 2 and taking the inverse cosine, we get
$\Rightarrow \dfrac{A}{2} = 41.4^\circ $ which corresponds to option (A);$ A = 82.8^\circ $.
Note
A prism with a refractive index dipped in some liquid is a convenient way used to measure the refractive index of a material (liquid or gas) by directing a light ray through a prism that is filled with the liquid or gas at minimum deviation experimentally. In our case, since we know the refractive index, we can find out the angle of the prism using the equation that we used in the solution.
In this solution we will be using the following formula,
$\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$ where $\mu $ is the refractive index, $A$ is the angle of the prism and ${\delta _m}$ is the angle of minimum deviation.
Complete step by step answer
The minimum deviation ${\delta _m}$ in a prism occurs when the entering angle and the exiting angle are the same. In this situation, the ray of light when inside the prism is parallel to the base of the prism. The angle of refraction inside the prism on either side of the prism is also the same in this situation.
If we are given that angle of minimum deviation and the angle of the prism, we can find the refractive index of the prism $\mu $ as:
$\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
Since we have been given in the question as${\delta _m} = A$, we can substitute it in the above equation and write
$\Rightarrow \mu = \dfrac{{sin\left( {\dfrac{{A + A}}{2}} \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
$\Rightarrow \mu = \dfrac{{sin\left( A \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
Since $\mu = 1.5$, we have
$\Rightarrow 1.5 = \dfrac{{sin\left( A \right)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
Now, we know that $\sin (A) = 2\sin (A/2)\cos (A/2)$ so we can write
$\Rightarrow 1.5 = \dfrac{{2\sin (A/2)\cos (A/2)}}{{sin\left( {\dfrac{A}{2}} \right)}}$
$\Rightarrow 1.5 = 2\cos (A/2)$
On dividing both sides by 2 and taking the inverse cosine, we get
$\Rightarrow \dfrac{A}{2} = 41.4^\circ $ which corresponds to option (A);$ A = 82.8^\circ $.
Note
A prism with a refractive index dipped in some liquid is a convenient way used to measure the refractive index of a material (liquid or gas) by directing a light ray through a prism that is filled with the liquid or gas at minimum deviation experimentally. In our case, since we know the refractive index, we can find out the angle of the prism using the equation that we used in the solution.
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