Name: Ray Diagram: Lenses
Uploaded: 2019-07-19
Duration: 6 min 35 s
Description: In this video, let us discuss how ray diagrams for lenses can be used to determine the size and location of a refracted image. Get in to know more!

Question

Deduce an expression for mirror equation. Draw necessary ray diagram.

Accepted Answer

Derivation of mirror formula for a concave mirror when it forms a real image. Consider an object AB placed on the principal axis beyond the centre of curvature C of a concave mirror of small aperture, as

Object distance, $B P = - u$

Image distance, $B' P = - v$

focal length, $F P = - f$

Radius of curvature, $C P = - R = - 2 f$

Now, $∆ A' B' C ~ ∆ A B C$

$∴ \frac{A' B'}{A B} = \frac{C B'}{B C} = \frac{C P - B' P}{B P - C P} = \frac{- R + v}{- u + R} . . . . . . . . . . (1)$

as $∠ A' P B' = ∠ A B P, t h e r e f o r e,$

$∆ A' B' P ≅ ∆ A B P$

consequently,

$\frac{A' B'}{A B} = \frac{B' P}{B P} = \frac{- v}{- u} = \frac{v}{u} . . . . . . . . . . . . . (2)$

From equation (1) and (2), we get,

$\frac{- R + u}{- u + R} = \frac{v}{u}$

or, $v R + u R = 2 u v$

dividing both sides by uvR we get,

$\frac{1}{u} + \frac{1}{v} = \frac{2}{R}$

But R=2f

$∴ \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

Deduce an expression for mirror equation. Draw necessary ray diagram.

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