TIR

A rod of glass μ=3/2 and of square of cross section is bent into the shape shown in A parallel beam of light falls perpendicularly on the plane flat surface A.Referring to the diagram,d is the width of the side and R is the radius of the inner semi-circle,Find the maximum value of Ratio of d/R so that all the light enetering the glass through the surface A emerge from the glass through B.

16 Answers

24
eureka123 ·

3
msp ·

y u r taking only one ray da

can u explain more on ur inequality

24
eureka123 ·

I am talking about the extreme ray through A which is incident at pt Q........becoz if this ray will get totally internally reflected then remaining will automatically will totally internally reflected.....

this is the extreme cond......[1][1]

3
msp ·

how did u conclude dat the ray striking at Q will be the extreme ray.

i cant understand dat thing .dats y i asked can u pls give me more explanation

11
Anirudh Narayanan ·

Machans, it will be the right extreme ray

"becoz if this ray will get totally internally reflected then remaining will automatically will totally internally reflected....."

Can u pls explain this statement , eureka? I hv difficulty in imagining it, da. [2]

24
eureka123 ·

wait a minute ..........i will draw a figure and the n u will know....

3
msp ·

yup got it eureka no need to draw.tx da.

This ray must be the extreme ray becos this ray have incident angle lesser than the other rays da.correct me if i was wrong.

11
Anirudh Narayanan ·

Ya got it. Good one eureka [4]

24
eureka123 ·

24
eureka123 ·

oops....didnt read ur post..........anyways u can make it out now....

3
msp ·

tx da.

11
Anirudh Narayanan ·

Thanx eureka [1]

3
msp ·

eureka but the incident angle will be 45 degrees as in the figure so it always undergoes tir

so no need to solve the question am i rite eureka

11
Anirudh Narayanan ·

for the right extreme ray, i is always 45°

24
eureka123 ·

why 45°??????????[11]

3
msp ·

consider the ray reflected at Q

wkd i=r and the rays are at 90 degrees therfore i=45

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