Question

A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and its is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
(a) must be less than 10 cm
(b) must be greater than 20 cm
(c) must not be greater than 20 cm
(d) must not be less than 10 cm.

Answer 1

Answer:

A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and its is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens

(a) must be less than 10 cm

(b) must be greater than 20 cm

(c) must not be greater than 20 cm✔

(d) must not be less than 10 cm.

Answer 2

(a) must be less than 10 cm

Explanation:

We have the relation between the distance of object 'o' and image 'i' from the lens of focal length 'f' as:

$\frac{1}{f} =\frac{1}{i} +\frac{1}{o}$

$\frac{1}{f} =\frac{1}{x} +\frac{1}{40-x}$

here x is the distance of the image from the lens.

$\Rightarrow x^2-40x+40f=0$

Now applying Shreedharacharya's rule to determine the roots, we've:

$x=\frac{-(-40)\pm\sqrt{(-40)^2-4(1\times40f)} }{2\times1}$

$x=\frac{1600\pm \sqrt{(1600-160f)}}{2}$

So, if the value under the root becomes negative we will get imaginary roots showing that no such value of x is practically possible.

Therefore:

$1600<160f$

$f>10\ cm$

TOPIC: Thin Lens Equation, Sridharacharya formula

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