Math, asked by crimsoncupid, 7 months ago

A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance M from it. Find value of M for which ray from p will emerge parallel.​

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Answered by queen4bad
1

Answer :

  • The value of m is 4/3.

Solution :

 \sf \: by \: applying \:  \frac{u2}{v}  -  \frac{u1}{u}  =  \frac{u2 - u1}{r}

  • Firstly at plane surfaces and then at curved surfaces.

 \sf \: for \: plane \: surfaces \: we \: get :

 \frac{1.5}{AI"}  -  \frac{1}{( - mR)}  =  \frac{1.5 - 1}{ \infin}  = 0

AI" = (1.5mR)

 \sf \: for \: the \: curved \: surfaces \: ,  \\ \sf \: \: since \: the \: final  \: image \: is \: formed \: at \: infinity \: ,  \\  \sf \: \: so \: we \: get \: ,

 \frac{1}{ \infin}  -  \frac{1.5}{( - 1.5mR + R)}  =  \frac{1 - 1.5}{ - R}

 \implies \frac{1.5}{(1.5m + 1)R}  =  \frac{0.5}{R}

 \implies \: 3 = 1.5m \:  + 1

 \implies \:  \frac{3}{2} m = 2

 \implies \: m =  \frac{4}{3}

Therefore, The value of m is 4/3.

Answered by XxBadCaptainxX
0

Answer:

Answer in attachment..

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