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Q 9.6 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^{\circ}. What is the refractive index of the material of the prism? The refracting angle of the prism is 60^{\circ}. If the prism is placed in water, predict the new angle of minimum deviation of a parallel beam of light.

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In Prism :

Prism angle (A) =  First Refraction Angle ( r _1) + Second refraction angle ( r _2)

also, Deviation angle (\delta) = incident angle(i) + emerging angle(e)  -  Prism angle (A) ..............(1)

the deviation angle is minimum when the incident angle(i) and an emerging angle(e) are the same. in other words

                                                         i=e ...........(2)

from (1) and (2)

\delta_{min} = 2i-A

i=\frac{\delta_{min} +A}{2}..........................(3)

We also have 

  r _1 = r_2 = r = \frac{A}{2}   .................(4)

Now applying snells law using equation (3) and (4)

                                     \mu _1sini=\mu _2sinr

                          1sin(\frac{\delta _{min}+A}{2})=\mu _2sin\frac{A}{2}

                            \mu _2=\frac{sin(\frac{\delta _{min}+A}{2})}{sin\frac{A}{2}}...................(5)

Given

                                     \\\delta _{min}= 40 \\A = 60

putting those values in (5) we get

                              \mu _2=\frac{sin(\frac{40+60}{2})}{sin\frac{60}{2}}=\frac{sin50}{sin30}=1.532

Hence the refractive index of the prism is 1.532.

Now when the prism is in the water.

 Applying Snell's law: 

                                         \mu _1sin(\frac{\delta _{min}+A}{2})=\mu _2sin\frac{A}{2}

                                        1.33sin(\frac{\delta _{min}+60}{2})=1.532sin\frac{60}{2}

                                       sin(\frac{\delta _{min}+60}{2})=\frac{1.532*0.5}{1.33}

                                      \frac{\delta _{min}+60}{2}=sin^{-1}\frac{1.532*0.5}{1.33}

                                      \delta _{min} =2sin^{-1}0.5759 - 60

                                  \delta _{min} =2*35.16- 60 =10.32^0

Hence  minimum angle of deviation inside water is 10.32 degree.

Posted by

Pankaj Sanodiya

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