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A point source of monochromatic light S is kept at the centre of trhe bottom of A CYLINDER  of radius 15cm.The cylinder contains water of refractive index 4/3 to a height of 7cm. Draw the ray diagram and calculate trhe area of water surface through which the light emerges in air.

Asked by vasturushi 2nd February 2018, 6:47 PM
Answered by Expert
if angle of incidence is equal to the critical angle,
no light is refracted back to air
the light will appear as if it is emerging from the circle with an angle theta from the vertical
let the radius of that circle be 'r', the area will therefore be, πr squared
refractive index of air with respect to water is 4/3
begin mathsize 12px style sin space straight theta equals fraction numerator 1 over denominator 4 over 3 end fraction
thus space straight theta space equals 48.76 to the power of straight o
by space geometry space of space the space figure comma space
straight r over AC equals space Tanθ
substiuiting space all space the space values space we space get space straight r equals space 8 cm comma
thus space area space of space the space circle space becomes space
straight pi cross times 8 squared equals 201 space cm squared end style

Answered by Expert 4th February 2018, 4:09 PM
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