A screen is placed 90 cm away from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Find the focal length of lens.

Question

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Answer 1

The image of the object can be located on the screen for two-position of convex lens such that u and v are exchanged.
The separation between two positions of the lens is, d= 20 cm
From the figure,
$u_{1} + v_{1} = 90 c m$
$v_{1} - u_{1} = 20 c m$
Solving,
$v_{1} = 55 c m, u_{1} = 35 c m$
From lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{5 5} - \frac{1}{- 3 5} = \frac{1}{f}$

$\Rightarrow \frac{1}{f} = \frac{1}{5 5} + \frac{1}{3 5} = \frac{5 5 \times 3 5}{9 0}$
$\Rightarrow f = 21.4 c m$

Hence, the correct option is $(B)$

Answer 2

The image of the object can be located on the screen for two-position of convex lens such that u and v are exchanged.

The separation between two positions of the lens is, d= 20 cm

From the figure,

u_{1} + v_{1} = 90 c m

v_{1} - u_{1} = 20 c m

Solving,

v_{1} = 55 c m, u_{1} = 35 c m

From lens formula,

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

\frac{1}{5 5} - \frac{1}{- 3 5} = \frac{1}{f}

\Rightarrow \frac{1}{f} = \frac{1}{5 5} + \frac{1}{3 5} = \frac{5 5 \times 3 5}{9 0}

\Rightarrow f = 21.4 c m

Hence, the correct option is

(B)