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Standard X

Physics

Magnification

An illuminate...

Question

An illuminated bulb and a screen are placed $90 c m$ apart. What is the focal length of the lens required to produce a real image twice the size of object, on the screen?

f = - 20 c m

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f = - 10 c m

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f = 20 c m

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f = 10 c m

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Solution

The correct option is C $f = 20 c m$
Given:
Distance between object and image, $d = 90 c m$
Magnitude of magnification, $| m | = 2$
Image is real

As the image is real, the lens must be a convex lens and it should be placed between object and screen.
Let the object distance be $- u$ .
Then the image distance will be $90 - u$ .

Using formula for magnification,
$m = \frac{v}{u}$
$- 2 = \frac{90 - u}{- u}$
New cartesian sign convention used here.
$u = 30 c m$

Object distance, $u = - 30 c m$
Image distance, $v = - 2 u = 60 c m$

By lens formula,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{60} - \frac{1}{(- 30)}$
$f = 20 c m$

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An illuminated bulb and a screen are placed 90 cm apart. What is the focal length of the lens required to produce a real image twice the size of object, on the screen?

An illuminated bulb and a screen are placed $90 c m$ apart. What is the focal length of the lens required to produce a real image twice the size of object, on the screen?