Home » Ask & Discuss » Physics. » Optics « Back to Discussion

Optics

New kid on the Block

Joined:	28 Mar 2009
Post:	25

7 Apr 2009 10:30:35 IST

0 People liked this

333

optics

None

(voilet)

Share this article on:

Comments (4)

saharsha kumar keshkar

Forum Expert

Joined: 7 Aug 2008

Posts: 1104

7 Apr 2009 10:36:51 IST

Like 0 people liked this

solving...wait..

Prakhar Banga

Blazing goIITian

Joined: 20 Dec 2008

Posts: 599

7 Apr 2009 10:38:01 IST

Like 1 people liked this

Applying the equation u2/v - u1 / u = u2 - u1 / R for first refraction.

1.5 / v + 1 / 2R = 0.5 / R

So v = infinite. This means that the rays become parallel after the first refraction.

So now they converge and meet at the focus of the sphere, i.e. R/2 from the second pole.

These rays then go through the left surface again.

1 / v + 1.5 / (3R / 2) = -1/R.

v = - R / 2.

So while viewing from air, the image appears to be at distance R / 2 from the left curved surface.

Sorry there was a mistake

1 / v + 1.5 / (3R / 2) = 0.5/R.

v = - 2R.

Saharsha is right.

saharsha kumar keshkar

Forum Expert

Joined: 7 Aug 2008

Posts: 1104

7 Apr 2009 10:59:28 IST

Like 4 people liked this

Prakhar Banga

Blazing goIITian

Joined: 20 Dec 2008

Posts: 599

7 Apr 2009 11:01:14 IST

Like 2 people liked this

In the above figure, when the ray after reflection meets the principal axis at a distance of R / 2 towards the right from the centre, and when the ray gets out in air and is extended back, it meets at the silvered surface.

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team