For a glass prism $(\mu =\sqrt{3} )$ the angle of minimum deviation is equal to the angle of the prism. Calculate the angle of the prism.

Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having refractive index $\mu = \sqrt{3} . $

Q: 1. For a glass prism $(\mu =\sqrt{3} )$ the angle of minimum deviation is equal to the angle of the prism. Calculate the angle of the prism. 2. Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having refractive index $\mu = \sqrt{3} . $

1. $\mu = \frac{\sin(\frac{\text{A} + \text{D}}{2})}{\sin\frac{\text{A}}{2}}$ $ = \frac{\sin(\frac{2\text{A}}{2})}{\sin\frac{\text{A}}{2}} =2 \cos\text{A}\sqrt{2} = \sqrt{3}$ $\therefore\text{A} = 60 ^{o}$ 2. $\mu = \sqrt{3} = \frac{1}{\sin\text{i}_{c}}$ $\therefore\sin\text{i}_{c} = \frac{1}{\sqrt{3}}\cong0.58$ Lies between $30^\circ$ and $45^\circ$ Hence, TIR takes place. Alternate Answer $\sin\text{C} = \frac{1}{\sqrt{3}}$ which is less than $\frac{1}{\sqrt{2}}$ $\therefore $ Angle of incidence $ > i_c$ $\therefore\text{TIR}$

Question

For a glass prism $(\mu =\sqrt{3} )$ the angle of minimum deviation is equal to the angle of the prism. Calculate the angle of the prism.
Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having refractive index $\mu = \sqrt{3} . $

CBSE OUTSIDE DELHI - SET 3 SOUTH 2016

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Solution

$\mu = \frac{\sin(\frac{\text{A} + \text{D}}{2})}{\sin\frac{\text{A}}{2}}$

$ = \frac{\sin(\frac{2\text{A}}{2})}{\sin\frac{\text{A}}{2}} =2 \cos\text{A}\sqrt{2} = \sqrt{3}$
$\therefore\text{A} = 60 ^{o}$

$\mu = \sqrt{3} = \frac{1}{\sin\text{i}_{c}}$

$\therefore\sin\text{i}_{c} = \frac{1}{\sqrt{3}}\cong0.58$
Lies between $30^\circ$ and $45^\circ$
Hence, TIR takes place.
Alternate Answer
$\sin\text{C} = \frac{1}{\sqrt{3}}$ which is less than $\frac{1}{\sqrt{2}}$
$\therefore $ Angle of incidence $ > i_c$
$\therefore\text{TIR}$

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