Question

a ray of light on passing through an equilateral glass prism, suffers a minimum deviation equal to the angle of prism .The value of refractive index of the material of the prism is_______​

Answer 1

Answer:

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Explanation:

Minimum deviation δ=2i−A

where A is prism angle A=60  

o

 

For minimum angle the angle of refraction should be half of prism angle.

So r=A/2=30  

o

 

now using the snell's law

n  

air

​  

sini=n  

glass

​  

sinr

sini=1.5sin30

i=48.6  

o

 , incidence angle

Minimum deviation δ=2i−A

δ=2×48.6−60=37.9  

Answer 2

$\mu=1.732$ is the refractive index of the material of the prism

Explanation:

The relation between the minimum deviation angle and the refractive index of the prism material is given as:

$\mu=\frac{\sin(\frac{A+\delta m}{2}) }{\sin(\frac{A}{2}) }$

where:

$A=$ top vertical angle of the prism formed through the intersection of the surfaces involved in the refraction.

$\delta m=$ minimum deviation angle

As given in the question the prism is an equilateral triangle, so we've:

$A=60^{\circ}$

$\delta m=A=60^{\circ}$ (given)

Now putting the values in the above eq.

$\mu=\sin(\frac{60+60}{2} )\div \sin(\frac{60}{2} )$

$\mu=1.732$

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TOPIC: minimum deviation angle

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