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Standard XII

Physics

Magnification by Thin Lenses

The distance ...

Question

The distance between an object and the screen is $100$ cm. A lens produces an image on the screen when placed at either of two positions $40$ cm apart. The power of the lens is approximately.

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Solution

The correct option is B $5$ D
Given, $D = 100 c m, x = 40 c m$
$∴ f = \frac{D^{2} - x^{2}}{4 D}$
$= \frac{(100)^{2} - (40)^{2}}{4 \times 100}$
$= \frac{(100 + 40) (100 - 40)}{400}$
$= \frac{140 \times 60}{400} = 21$ cm
We know that $P = \frac{100}{f (c m)} = \frac{100}{21} ≃ 5 D$ .

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The distance between an object and the screen is 100cm. A lens produces an image on the screen when placed at either of two positions 40cm apart. The power of the lens is approximately.

The correct option is B 5DGiven, D=100cm,x=40cm∴f=D2−x24D=(100)2−(40)24×100=(100+40)(100−40)400=140×60400=21cmWe know that P=100f(cm)=10021≃5D.

The distance between an object and the screen is $100$ cm. A lens produces an image on the screen when placed at either of two positions $40$ cm apart. The power of the lens is approximately.

The correct option is B $5$ D
Given, $D = 100 c m, x = 40 c m$
$∴ f = \frac{D^{2} - x^{2}}{4 D}$
$= \frac{(100)^{2} - (40)^{2}}{4 \times 100}$
$= \frac{(100 + 40) (100 - 40)}{400}$
$= \frac{140 \times 60}{400} = 21$ cm
We know that $P = \frac{100}{f (c m)} = \frac{100}{21} ≃ 5 D$ .