A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used.

A converging lens is kept coaxially in contact with a diverging lens - both the lenses being of equal focal length. What is the focal length of the combination?

Q: 1. A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used. 2. A converging lens is kept coaxially in contact with a diverging lens - both the lenses being of equal focal length. What is the focal length of the combination?

1. For first position of the lens, we have $\frac{1}{\text{f}} = \frac{1}{\text{y}} -\frac{1}{(-\text{x})}$ For second position of the lens, we have $\frac{1}{\text{f}} = \frac{1}{\text{y - 20 }} -\frac{1}{(-(\text{x + 20 )})}$ $\frac{1}{\text{y}} +\frac{1}{\text{x}} =\frac{1}{(\text{y - 20)}} + \frac{1}{(\text{x + 20)}}$ $\frac{\text{x + y }}{\text{xy}} =\frac{(\text{x + 20)} + (\text{y - 20 )}}{(\text{y - 20 )}(\text{x + 20 )}}$ $\therefore\text{xy} = (\text{y - 20) }(\text{x + 20})$ $ = \text{xy} - 20 \text{x +20}\text{y - 400 }$ $\therefore\text{x - y } = - 20 $ Also , x + y =100 $\therefore$ x = 40 cm and y = 60 cm $\therefore\frac{1}{\text{f}} = \frac{1}{60} - \frac{1}{-40} =\frac{2+3}{120} =\frac{5}{120}$ $\therefore\text{f} = 24 \text{cm}$ Alternate Answer We have $\text{f} = \frac{\text{D}^{2} - \text{d}^{2}}{4\text{D}}$ $ = \frac{100^{2} - 20 ^{2}}{4\times100}$ $ =\frac{120\times80}{400}$ = 24 cm. Alternate Answer For the two positions of the lens, the values of the magnitudes of u and v, get interchanged. Hence, $| \text{u + v}| = 100$ $| \text{u - v}| = 20 , $ This gives $| \text{u}| = 60 |\text{v}| = 40 $ $\therefore\text{f} = 24 \text{ cm}$ Alternate Answer $2\text{x} + 20 = 100$ $\therefore\text{x} = 40 \text{ cm}$ For lens at position $L_1; \text{u} = -\text{x} =- 40 \text{cm} $ $\text{v} = 20 + 40 = 60\text{ cm}$ This gives f = 24 cm. 2. For combination of two lenses in contact. Net Power of combination, $P = P_1 + P_2$ $P_1 = +P , P_2 = -P$ So P= 0 and F= infinite Alternate Answer $\frac{1}{\text{F}} = \frac{1}{\text{f}_{1}} + \frac{1}{\text{f}_{2}}$ $ = \frac{1}{\text{f}} + (\frac{-1}{\text{f}}) = 0 $ F = infinite.

Question

A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used.
A converging lens is kept coaxially in contact with a diverging lens - both the lenses being of equal focal length. What is the focal length of the combination?

CBSE OUTSIDE DELHI - SET 1 NORTH 2016

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Solution

For first position of the lens, we have

$\frac{1}{\text{f}} = \frac{1}{\text{y}} -\frac{1}{(-\text{x})}$

For second position of the lens, we have
$\frac{1}{\text{f}} = \frac{1}{\text{y - 20 }} -\frac{1}{(-(\text{x + 20 )})}$
$\frac{1}{\text{y}} +\frac{1}{\text{x}} =\frac{1}{(\text{y - 20)}} + \frac{1}{(\text{x + 20)}}$
$\frac{\text{x + y }}{\text{xy}} =\frac{(\text{x + 20)} + (\text{y - 20 )}}{(\text{y - 20 )}(\text{x + 20 )}}$
$\therefore\text{xy} = (\text{y - 20) }(\text{x + 20})$
$ = \text{xy} - 20 \text{x +20}\text{y - 400 }$
$\therefore\text{x - y } = - 20 $
Also , x + y =100
$\therefore$ x = 40 cm
and y = 60 cm
$\therefore\frac{1}{\text{f}} = \frac{1}{60} - \frac{1}{-40} =\frac{2+3}{120} =\frac{5}{120}$
$\therefore\text{f} = 24 \text{cm}$
Alternate Answer
We have
$\text{f} = \frac{\text{D}^{2} - \text{d}^{2}}{4\text{D}}$
$ = \frac{100^{2} - 20 ^{2}}{4\times100}$
$ =\frac{120\times80}{400}$
= 24 cm.
Alternate Answer
For the two positions of the lens, the values of the magnitudes of u and v, get interchanged.
Hence, $| \text{u + v}| = 100$
$| \text{u - v}| = 20 , $
This gives $| \text{u}| = 60 |\text{v}| = 40 $
$\therefore\text{f} = 24 \text{ cm}$
Alternate Answer

$2\text{x} + 20 = 100$
$\therefore\text{x} = 40 \text{ cm}$
For lens at position $L_1; \text{u} = -\text{x} =- 40 \text{cm} $
$\text{v} = 20 + 40 = 60\text{ cm}$
This gives f = 24 cm.

For combination of two lenses in contact.

Net Power of combination,
$P = P_1 + P_2$
$P_1 = +P , P_2 = -P$
So P= 0 and F= infinite
Alternate Answer
$\frac{1}{\text{F}} = \frac{1}{\text{f}_{1}} + \frac{1}{\text{f}_{2}}$
$ = \frac{1}{\text{f}} + (\frac{-1}{\text{f}}) = 0 $
F = infinite.

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