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A right angled prism of apex angle 40 and r. i. 1.5 is located in front of vertical plane mirror as shown in fig.A horizontal ray of light is falling on the prism. Find the total deviation produced in the light ray at it emerges second time from the prism:

Option: 1

8cw


Option: 2

6cw


Option: 3

180cw


Option: 4

176cw


Answers (1)

best_answer

As we learn

Deviation from thin prism -

\delta = \left ( \mu -1 \right )A

- wherein

Applicable when A is very small

 ( i.e. thin prism )

 

 Deviation produced by prism is 

\delta _{1}=(\mu -1)A=2cw

Angle of incidence of mirror is \delta _{1}, so deviation produced by mirror is 

\delta _{2}=\pi -2\delta _{1}=176^{0}cw

deviation produced by the prism for second refraction is 

\delta _{3}=2^{0}Acw

Net deviation is 1760 cw

 

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Gunjita

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