CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Lens Maker's Formula

The magnifica...

Question

The magnifications produced by a convex lens for two different of an object are $m_{1}$ and $m_{2}$ respectively $(m_{1} > m_{2})$ . If 'd' is the distance of separation between the two positions of the object then the focal length of the lens is

A

\sqrt{m_{1} m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{d}{\sqrt{m_{1} m_{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

\frac{d m_{1} m_{2}}{m_{1} - m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

\frac{d}{m_{1} - m_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{d}{m_{1} - m_{2}}$
Let us consider $m_{1} > m_{2}$
$\frac{v}{u} = m_{1}$
For case ii, $m_{2} = \frac{u}{v}$
$m_{1} m_{2} = 1$
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{u} + \frac{1}{m_{1} u} = \frac{1}{f}$
$u = \frac{f (1 + m_{1})}{m_{1}}$
$u = m_{2} v$
$v = \frac{f (1 + m_{2})}{m_{2}}$
$d = v - u = f (\frac{1 + m_{2}}{m_{2}} - \frac{1 + m_{1}}{m_{1}})$
$d = f (\frac{m_{1} + m_{1} m_{2} - m_{2} - m_{1} m_{2}}{m_{1} m_{2}})$
$f = \frac{m_{1} m_{2}}{m_{1} - m_{2}}$
$m_{1} m_{2} = 1$
$f = \frac{d}{m_{1} - m_{2}}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A lens of focal length

^{'} f^{'}

is placed in between an object and screen at a distance

^{'} D^{'}

. The lens forms two real images of object on the screen for two of its different positions, a distance

^{'} x^{'}

apart. The two real images have magnifications

m_{1}

m_{2}

(m_{1} > m_{2})

Q. In the displacement method, a convex lens is placed in between an object and a screen.If the magnification in the two positions are,

m_{1}

m_{2}

m_{1} > m_{2}

) and the distance between the two positions of the lens is x, the focal length of the lens is

Q. In the displacement method, a convex lens is placed in between an object and a screen. If the magnifications in the two positions are

m_{1}

m_{2}

and the displacement of the lens between the two positions is x, then the focal length of the lens is

Q. The lateral magnification of the lens with an object located at two different positions

u_{1}

u_{2}

m_{1}

m_{2}

, respectively. Then the focal length of the lens is :

Q. An object and screen are fixed at a distance

d

apart. When a lens of focal length

f

is moved between the object and the screen, sharp images of the object are formed on the screen for two potions of the lens. The magnifications produced at these two positions are

M_{1}

M_{2}

.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Adaptive Q14

PHYSICS

Watch in App

explore_more_icon

Explore more

Lens Maker's Formula

Standard X Physics

Join BYJU'S Learning Program

CrossIcon