Name: Combination of a Lens and a Mirror
Uploaded: 2019-07-19
Duration: 7 min 19 s
Description: What would be the result of the combination of lenses and mirrors? Watch this video to learn about image formation in a combination of lenses and mirror.

Question

A convex lens is placed between an object and a screen. For two positions of the lens, real images are produced on the screen. If the lengths of the two images be

I_{1}

and

I_{2}

and that of the object be

O

, prove that

O = \sqrt{I_{1} I_{2}}

Accepted Answer

Let $D$ be the distance between object and screen. Let us mark the position of lens $I_{2}$ . Then let us move the lens away from the object to get a diminished image. Let this position of lens be $O$ . let $x$ be the distance between the lens positions $x_{1}$ and $x_{2}$ .
Let $v$ be the distance between image and lens and $u$ be the distance between object and lens.

Using lens equation, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

From figure, $v = D - u$ , then $\frac{1}{D - u} - \frac{1}{u} = \frac{1}{f}$ $\Rightarrow u^{2} - D u + f D = 0$

Solution for above equation is $u = \frac{D \pm \sqrt{D^{2} - 4 f D}}{2}$ .

When, $D = 4 f$ , we get only one position of lens to get image.

Hence, when $u_{1} = \frac{D - \sqrt{D^{2} - 4 f D}}{2}$ , then $v_{1} = \frac{D + \sqrt{D^{2} - 4 f D}}{2}$ and

when $u_{2} = \frac{D + \sqrt{D^{2} - 4 f D}}{2}$ , then $v_{2} = \frac{D - \sqrt{D^{2} - 4 f D}}{2}$

Now from magnification formula $m = \frac{I}{O} = \frac{- v}{u}$ .

$\frac{I_{1}}{O} = \frac{- v_{1}}{u} = - \frac{\frac{D + \sqrt{D^{2} - 4 f D}}{2}}{\frac{D - \sqrt{D^{2} - 4 f D}}{2}} = - \frac{D + \sqrt{D^{2} - 4 f D}}{D - \sqrt{D^{2} - 4 f D}} . . . (i)$

$\frac{I_{2}}{O} = \frac{- v_{2}}{u} = - \frac{\frac{D - \sqrt{D^{2} - 4 f D}}{2}}{\frac{D + \sqrt{D^{2} - 4 f D}}{2}} = - \frac{D - \sqrt{D^{2} - 4 f D}}{D + \sqrt{D^{2} - 4 f D}} . . . (i i)$

Multiplying both equation,

$\frac{I_{1}}{O} \times \frac{I_{2}}{O} = (- \frac{D + \sqrt{D^{2} - 4 f D}}{D - \sqrt{D^{2} - 4 f D}}) \times (- \frac{D - \sqrt{D^{2} - 4 f D}}{D + \sqrt{D^{2} - 4 f D}}) = 1$

Therefore, $O = \sqrt{I_{1} I_{2}}$ .

A convex lens is placed between an object and a screen. For two positions of the lens, real images are produced on the screen. If the lengths of the two images be $I_{1}$ and $I_{2}$ and that of the object be $O$ , prove that $O = \sqrt{I_{1} I_{2}}$

Important Questions on Refraction at Spherical Surfaces and Spherical Lens

Learn and Prepare for any exam you want !

A convex lens is placed between an object and a screen. For two positions of the lens, real images are produced on the screen. If the lengths of the two images be I1 and I2 and that of the object be O, prove that O=I1I2

Important Questions on Refraction at Spherical Surfaces and Spherical Lens

Learn and Prepare for any exam you want !

A convex lens is placed between an object and a screen. For two positions of the lens, real images are produced on the screen. If the lengths of the two images be $I_{1}$ and $I_{2}$ and that of the object be $O$ , prove that $O = \sqrt{I_{1} I_{2}}$