Test Index

NCERT Class XII Chapter
Ray Optics and Optical Instruments
Questions With Solutions

© examsnet.com
Question : 6 of 38
Marks: +1, -0
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Solution:  
When the light beam is incident from air on to the glass prism, the angle of minimum deviation is 40°.
Refractive index of glass w.r.t. air.
 aμg\,{}^{a}\mu_g = sin⁡(A+δm)2sin⁡A2\frac{ \frac{ \sin(A+\delta_m) }{ 2 } }{ \frac{ \sin A }{ 2 } }
 aμg\,{}^{a}\mu_g = sin⁡(60∘+40∘)2sin⁡60∘2\frac{ \frac{ \sin(60^{\circ}+40^{\circ}) }{ 2 } }{ \frac{ \sin 60^{\circ} }{ 2 } } = sin⁡50∘sin⁡30∘\frac{ \sin 50^{\circ} }{ \sin 30^{\circ} } = 0.7660.50\frac{0.766}{0.50} = 1.532
Now the prism is placed in water, new angle of minimum deviation can be calculated.
 ωμg\,{}^{\omega}\mu_g = sin⁡(A+δm′)2sin⁡A2\frac{ \frac{ \sin(A+\delta_m') }{ 2 } }{ \frac{ \sin A }{ 2 } }
 aμg aμω\frac{\,{}^{a}\mu_g}{\,{}^{a}\mu_{\omega}} = sin⁡(60+δm′)2sin⁡30∘\frac{ \frac{ \sin(60+\delta_m') }{ 2 } }{ \sin 30^{\circ} }
sin (30∘+δm′2)\left(30^{\circ} + \frac{\delta_m'}{2}\right) = 12[0.5321.33]\frac{1}{2} \left[ \frac{0.532}{1.33} \right] = 0.5759
30° + δm′2\frac{\delta_m'}{2} = sin⁡−1\sin^{-1} (0.5759) or 30° + δm′2\frac{\delta_m'}{2} = 35° 10'
New angle of deviation δm′\delta'_m = 10° 20′
© examsnet.com
Go to Question:
Prev Question
Next Question