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

An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation

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

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Accepted Answer

Ray diagram to show the formation of a virtual image by concave mirror:

Derivation of mirror equation using the above ray diagram:

$∆ C A^{'} B^{'}$ and $∆ C A B$ are similar triangles. Hence, we can write:

$\frac{A^{'} B^{'}}{A B} = \frac{C B^{'}}{C B} = \frac{C P + P B^{'}}{C P - B P}$ .......(i)

$D N$ is equal to $A B$ and is a normal on the principal axis.

$∆ F D N$ and $∆ F A^{'} B^{'}$ are similar triangles. The incident rays are considered to be paraxial rays hence, $D and P$ are very close. We can write:

$\frac{A^{'} B^{'}}{A B} = \frac{F B^{'}}{F D} = \frac{F P + P B^{'}}{F D} = \frac{F P + P B^{'}}{F P}$ .....(ii)

Using the above equations and substituting the values according to the sign convention we get:

$\frac{C P + P B^{'}}{C P - B P} =$ $\frac{F P + P B^{'}}{F P}$

$\frac{- 2 f + v}{- 2 f + u} = \frac{- f + v}{- f}$

$\Rightarrow (- 2 f^{2} - f v = 2 f^{2} - f u - 2 v f = u v)$

$\Rightarrow (f u + f v = u v f)$

On dividing both sides by $u v f$ we get:

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

The above equation is the required mirror equation.

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