IIT JEE , AIEEE Forum - Physics , Optics - physics-optics

iwanion

a circular disc of diameter d lies horizontally inside a metallic hemispherical bowl of radius a. the disc is just visible to an eye looking over the edge. the bowl is now filled with a liquid of refrative index

.now, the whole of the disc is just visible to the eye in the same position.show that d=2a(

²-1)/(

²+1)

aditya_arora04

Where did you get this question. Please tell. I think something is wrong with this question

iwanion

no the question is correct

iwanion

i solved it once but now the solution had skipped my mind.and i do not have the solution

krishna.gopal

The depth of the disc below the flat surface of hemispherical bowl will be
h=sqrt(a^2-d^2)/2
So tan(i)=(a+d)/(2h)
tan(r)=(a-d)/(2h)
Solve these to sin(i) and sin(r)
and put sin(i)=

sin(r)

You will get the desired relation

Enjoy.....

iwanion

i think u do not know anything. this question i did once but now i had forgotten the way. as soon i will remember it i will also help you to solve such questions

krishna.gopal

Well friends i am sorry for my previous post where i have taken a as diameter of the hemispherical bowl by mistake while it was given that a is radius of the disc. Make this change and proceed on the same path and you will get the answer.

Let h be the height of the brim of the hemisphere from the disc

applying pythagorus theorm

(radius of hemisphere)²=h²+ (radius of disc)²

h=sqrt(a²-d²/4)

Now if you make the ray diagram

tan (i) = (a+d/2)/h

tan(r) = (a-d/2)/h

Use h as derived above and identity that sin²(x)=tan²(x)/(1+tan²(x))

we get sin²(i)=(a+d/2)²/(a²+d²/4+ad+a²-d²/4)=(a+d/2)²/(2a²+ad) = (a+d/2)/(2a)simillarly

sin²(r)=(a-d/2)²/(a²+d²/4-ad+a²-d²/4)=(a-d/2)²/(2a^2-ad) = (a-d/2)/(2a)

Thus

²/1 = sin²(i)/sin²(r) = (a+d/2)/(a-d/2)

So

(

²-1)/(

²+1)=d/(2a)

Or d= (2a)*(

²-1)/(

²+1)

I think this solves the purpose. Enjoy...........

Ask & Discuss Questions with Community & Experts