1.

A narrow beam of light is incident on a glass plate of refractive index 1.6. It makes an angle 53^@ with normal to the interface. Find the lateral shift of the beam at the point of emergence, if thickness of the plate is 30 mm. (Take sin53^@ = 0.8.)

Answer»

9.023 mm
15.52 mm
13.53 cm
13.53 mm

Solution :For incident ray,
`n_1sintheta_1=n_2sin theta_2`
(1) `sin53^@=1.6sintheta_2`
0.7986=1.6`sintheta_2`
`thereforesin theta_2=(0.7986)/(1.6)=0.4991 ~~0.5 " rad "=30^@`
Now, LATERAL shift,
`x=(TSIN(theta_1-theta_2))/(costheta_2)`
`=(30sin(53^@-30^@))/(COS30^@)=(30sin23^@)/(cos30^@)`
`=(30xx0.3907)/(0.866)=13.53` mm


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