IIT JEE , AIEEE Forum - Physics , Optics

kislay

a glass prism in the shape of quater cylinder lies on horzontal table. a uniform horizontal light beam falls on its vertical plane surface ,as shown in the figure.if the radius of the cylinder is R=5cm and the refractive index of tglas is n=1.5 ,where on the tavle beyonfd the cylinder will a path of light be fopund??

ans-->1.71cm

harsha_27

let the critical angle be @

now from the figure,

cos@ = R/(R+x)

=>R + x = R/cos@

=> x = (R/cos@) - R

= R(sec@ - 1 )

= R[(3/

5) - 1] {since sin@ = 2/3}

= 5[(3/

5) - 1]

= 3

5 - 5

= 1.7082039324993690892275210061938 cms

1.71 cms

kislay

how u have tken @ as the the critical angle???????
i mean light is incudent on other suface at critical angle, how????

bvsatyaram

The light remains as parallel beam after the first refraction...
So, this parallel beam converges at the focal point on the other side of the prism. So, using refraction at curved surface formula find focal length. The beam converges at the focal point of this prism.

harsha_27

that is the xtreme case of refraction yaar........after that angle T.I.R. takes place......means that ray forms the boundary of light falling on the table.....means that is the last ray to reach the table......got it?

kislay

got that.... cheers

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