Answer
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Hint: We need to note the criteria for minimum deviation to occur in an equilateral prism. As we know for minimum deviation to occur, the angle of refraction of the light should be equal to half of the angle of the prism. Then using Snell’s law we can calculate the angle of incidence. By substituting the angle of incidence in the formula, we get the minimum deviation.
Complete step by step answer:
The rays go through minimum deviation only when they pass symmetrically through the prism.
Minimum deviation $\delta =2i-A$
An equilateral prism has a prism angle A$={{60}^{\circ }}$
For minimum deviation, the angle of refraction should be half of the prism angle.
So $r=\dfrac{A}{2}={{30}^{\circ }}$
Now by using Snell’s law
${{n}_{a}}\sin i={{n}_{g}}\sin r$ (Where ${{n}_{a}}$is refractive index in air and ${{n}_{g}}$ is refractive index in glass)
$\sin i=1.5\sin 30$
$i={{48.6}^{\circ }}$
Minimum deviation $\delta =2\times 48.6-60={{37.2}^{\circ }}={{37}^{\circ }}10'$
Hence the correct answer is option C.
Note:
Angle of prism is the angle made by the two sides of the prism through which the incident light enters and the emergent light exits. In a prism, with increase in the angle of incidence, the angle of deviation decreases. This happens to a particular angle only. The position of the angle of incidence where the angle of deviation in a prism is minimum is called the Minimum Deviation Position of the prism and that very deviation angle is known as the Minimum Angle of Deviation. It is also responsible for phenomena like halos and sundogs, produced by the deviation of sunlight by the hexagonal ice crystals in the air bending light.
Complete step by step answer:
The rays go through minimum deviation only when they pass symmetrically through the prism.
![seo images](https://www.vedantu.com/question-sets/d8c7ac6b-f151-447e-b66f-a53f01857ad93362838090489847557.png)
Minimum deviation $\delta =2i-A$
An equilateral prism has a prism angle A$={{60}^{\circ }}$
For minimum deviation, the angle of refraction should be half of the prism angle.
So $r=\dfrac{A}{2}={{30}^{\circ }}$
Now by using Snell’s law
${{n}_{a}}\sin i={{n}_{g}}\sin r$ (Where ${{n}_{a}}$is refractive index in air and ${{n}_{g}}$ is refractive index in glass)
$\sin i=1.5\sin 30$
$i={{48.6}^{\circ }}$
Minimum deviation $\delta =2\times 48.6-60={{37.2}^{\circ }}={{37}^{\circ }}10'$
Hence the correct answer is option C.
Note:
Angle of prism is the angle made by the two sides of the prism through which the incident light enters and the emergent light exits. In a prism, with increase in the angle of incidence, the angle of deviation decreases. This happens to a particular angle only. The position of the angle of incidence where the angle of deviation in a prism is minimum is called the Minimum Deviation Position of the prism and that very deviation angle is known as the Minimum Angle of Deviation. It is also responsible for phenomena like halos and sundogs, produced by the deviation of sunlight by the hexagonal ice crystals in the air bending light.
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