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A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.?
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A ray of light passing through a prism having refractive index √2 suff...
Solution:
Given:
Refractive index of prism, $\mu = \sqrt{2}$
Angle of incidence, $i = 2r$
Let us first draw a diagram of the given situation:
Minimum Deviation:
For a given prism, the angle of incidence and angle of emergence for which the deviation produced is minimum is called the angle of minimum deviation. Let this angle be $D_m$.
We know that for a prism, angle of deviation is given by:
$$A = (μ − 1) (i + e − A)$$
Where, $i$ is the angle of incidence, $e$ is the angle of emergence and $A$ is the angle of prism.
For minimum deviation, $\frac{dA}{de} = 0$
Differentiating the above equation with respect to $e$, we get:
$$\frac{dA}{de} = \frac{(μ-1)}{2} \left(\frac{\sin i}{\sin(\frac{A+e}{2})} - \frac{\sin e}{\sin(\frac{A+e}{2})} - \cos(\frac{A+e}{2})\right) = 0$$
Simplifying the above equation, we get:
$$\sin i = (μ − 1) \sin \left(\frac{A}{2}\right)$$
Angle of Deviation:
Using Snell's Law, we can write:
$$\frac{\sin i}{\sin r} = \mu$$
As given, $i = 2r$
Substituting this value in the above equation, we get:
$$\frac{\sin 2r}{\sin r} = \sqrt{2}$$
Simplifying the above equation, we get:
$$2\cos r = \sqrt{2}$$
$$\cos r = \frac{\sqrt{2}}{2}$$
$$r = 45º$$
Therefore, $i = 90º$
Angle of Prism:
Using the relation $A = (μ − 1) (i + e − A)$, we can write:
$$A = (μ − 1) (i + e -A)$$
Substituting the values of $μ$, $i$ and $e$ in the above equation, we get:
$$A = (\sqrt{2}-1)(90º+45º-A)$$
Simplifying the above equation, we get:
$$A = 60º$$
Therefore, the angle of prism is $60º$.
Given:
Refractive index of prism, $\mu = \sqrt{2}$
Angle of incidence, $i = 2r$
Let us first draw a diagram of the given situation:
Minimum Deviation:
For a given prism, the angle of incidence and angle of emergence for which the deviation produced is minimum is called the angle of minimum deviation. Let this angle be $D_m$.
We know that for a prism, angle of deviation is given by:
$$A = (μ − 1) (i + e − A)$$
Where, $i$ is the angle of incidence, $e$ is the angle of emergence and $A$ is the angle of prism.
For minimum deviation, $\frac{dA}{de} = 0$
Differentiating the above equation with respect to $e$, we get:
$$\frac{dA}{de} = \frac{(μ-1)}{2} \left(\frac{\sin i}{\sin(\frac{A+e}{2})} - \frac{\sin e}{\sin(\frac{A+e}{2})} - \cos(\frac{A+e}{2})\right) = 0$$
Simplifying the above equation, we get:
$$\sin i = (μ − 1) \sin \left(\frac{A}{2}\right)$$
Angle of Deviation:
Using Snell's Law, we can write:
$$\frac{\sin i}{\sin r} = \mu$$
As given, $i = 2r$
Substituting this value in the above equation, we get:
$$\frac{\sin 2r}{\sin r} = \sqrt{2}$$
Simplifying the above equation, we get:
$$2\cos r = \sqrt{2}$$
$$\cos r = \frac{\sqrt{2}}{2}$$
$$r = 45º$$
Therefore, $i = 90º$
Angle of Prism:
Using the relation $A = (μ − 1) (i + e − A)$, we can write:
$$A = (μ − 1) (i + e -A)$$
Substituting the values of $μ$, $i$ and $e$ in the above equation, we get:
$$A = (\sqrt{2}-1)(90º+45º-A)$$
Simplifying the above equation, we get:
$$A = 60º$$
Therefore, the angle of prism is $60º$.
Community Answer
A ray of light passing through a prism having refractive index √2 suff...
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A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.?
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A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.?.
A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ray of light passing through a prism having refractive index √2 suffers minimum deviation. It is found that angle of incidence is double the Angle of refraction within the prism. Find angle of the prism.?.
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