Find the position of the image formed by the lens combination given in the figure

Open in App

Solution

For first lens

$f_{1} = + 10 c m$

$u_{1} = - 30 c m$

Using lens formula

$\frac{1}{v_{1}} - \frac{1}{u_{1}} = \frac{1}{f_{1}}$

$\frac{1}{v_{1}} = \frac{1}{f_{1}} + \frac{1}{u_{1}}$

$v_{1} = + 15 c m$

For the first lens image is formed at a distance of $15 c m$ from the right of it. This image behaves as object for second lens.

For second lens

$f_{2} = - 10 c m$

$u_{2} = 15 - 5 = + 10 c m$

$\frac{1}{v_{2}} = \frac{1}{f_{2}} + \frac{1}{u_{2}}$

$\frac{1}{v_{2}} = \frac{- 1}{10} + \frac{1}{10}$

$v_{2} = \infty$

Real image is formed by the second lens at infinite distance. This image will acts as an object for third lens.

For third lens

$f_{3} = + 30 c m$

$u_{3} = \infty$

$\frac{1}{v_{3}} = \frac{1}{f_{3}} + \frac{1}{u_{3}}$

$\frac{1}{v_{3}} = \frac{1}{30} + \frac{1}{\infty}$

$v_{3} = + 30 c m$

So, the final image is formed at distance of 30 cm to the right of third lens.

Suggest Corrections

Similar questions

(f_{1}, f_{2}

)

(f_{1} = + 10 c m, f_{2} = - 10 c m

f_{3} = + 30 c m)

45^{\circ}

O

P

^{'} L^{'}

^{'} M^{'}

^{'} O^{'}

^{'} I^{'}

^{'} O^{'}

Join BYJU'S Learning Program