Question

Write derivation of lens maker formula for concave lens.

Answer 1

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Answer 2

Consider a concave lens. Let ‘P’ be the optical centre or pole of the lens. ‘F’ and ‘C’ is the focus and centre of curvature of the lens respectively. Let us place an object XY in between the focus and the centre of curvature of the concave lens. Now draw a line parallel to the principle axis from the object. The point which touch the lens is marked as M. Since it is a concave lens, the light rays coming from the object will pass through the lens and diverge away from the focus. Draw another line from X to P and extend the divergent line to the focus. The point of intersection will be the image of the object, X’Y’.

The image is virtual, erect, diminished and is in between focus and pole. It is on the same side where the object is placed.    

Let –f be distance from the focus and the pole, -u be the distance from pole and the object,

-v be the distance from image and the pole.  

The negative sign indicates the image is on the left-hand side of the lens.

Consider the triangle XYP and triangle X’Y’P. XY is parallel to X’Y’ and angle Y and angle Y’ are right angled.  

From the fig (2)

$\frac{XY}{YP}=\frac{X'Y}{'Y'P}$  

Cross multiplying, we get,

$\frac{X'Y'}{XY}=\frac{Y'P}{YP}$  

= -v-u = vu ........................... (1)

Consider the second triangle FMP and FX’Y’. here MP is parallel to X’Y’ and both the angles P and Y’ are right angled.

$\frac{MP}{PF}=\frac{X'Y'}{Y'F}$  

Since MP = XY we can write $\frac{XY}{PF}= \frac{X'Y'}{Y'F}$

$\frac{X'Y'}{XY}= \frac{Y'F}{PF}$  

From the figure (3)

$\frac{X'Y'}{XY}= \frac{Y'F}{PF} = \frac{-f-(-v)}{-f} = \frac vu$  

From eqn. (1)  

$\frac{X'Y'}{XY} = \frac vu = \frac{f-v}f$

On simplification,

$vf = u(f-v)\\\\ vf = uf-uv\\ \\uv = f(u-v)$

$\frac{1}{f} = \frac{u-v}{uv}$

$\frac{u}{uv} - \frac{v}{uv}$

$\frac1f = \frac1u - \frac1v$