Question

An object of height 4 cm is placed 40 cm in front of a concave mirror of focal length 20cm. At what distance from the mirror should a screen be placed , so that a sharp image can be obtained? Find the size and nature of the image.

Answer 1

Explanation:

Given : Mirror type: Concave

Height of object (h) :(+) 4cm

Distance of object (u) : (-) 40cm

Focal length (f) : (-) 20cm

To find : Distance of image (v)

Size of image

Nature of image

Solution : By mirror formula,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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

$\frac{1}{v} = \frac{1}{( - 20)} - \frac{1}{( - 40)}$

$\frac{1}{v} = \frac{1}{ - 20} + \frac{1}{40}$

$\frac{1}{v} = \frac{ - 2 + 1}{40}$

$\frac{1}{v} = - \frac{1}{40}$

$v = - 40cm$

Size : i.e. Height of image (p)

$\frac{p}{h} = \frac{v}{u}$

$p = \frac{vh}{u}$

$p = \frac{( - 40)(4)}{ (- 40)}$

$p = ( + )4cm$

Therefore the screen should be placed at a distance of -40cm to get a sharp image of object and size of image is same as the size of object and (+) sign in (p) indicates the image is virtual and erect.

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