Question

An object of size 7.0 cm in length is placed at 27 cm in front of a concave minor of focal
length 18cm. At what distance from the mirror should a screen be a placed, so that a sharp
focused image can be obtained? Find the size and the nature of the image.​

Answer 1

Given :

• Size of object = 7 cm

• Distance of object from pole of mirror = 27 cm

• Focal length of concave mirror = 18 cm

To find:

• distance from the mirror should a screen be a placed, so that a sharp focused image can be obtained

• size and the nature of the image.

Formula used :

$\sf \large \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\sf \large m = - \dfrac{v}{u} = \dfrac{h_i}{h _o }$

Where :-

• v = Distance of image from pole of mirror
• u = Distance of object from pole of mirror
• hi = height of image
• ho = height of object
• m = magnification
• f = focal length

Solution :

To find distance of sharped imaged from the mirror, we will use the following formula :-

$\sf \implies \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

Put :-

• u = -27
• f = -18

$\sf \implies \dfrac{1}{v} - \dfrac{1}{27} = - \dfrac{1}{18}$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{27} - \dfrac{1}{18}$

LCM of 27 and 18 = 54

$\sf \implies \dfrac{1}{v} = \dfrac{2 - 3}{54}$

$\sf \implies \dfrac{1}{v} = \dfrac{ -1 }{54}$

$\sf \implies{v} = { - }{54}$

Distance of image from pole of mirror = 54 cm (behind object)

Now we will find nature of image.

$\sf m = - \dfrac{v}{u} = \dfrac{h_i}{h _o }$

$\sf \implies m = - \: \dfrac{ - 54}{ - 27} = - 2$

Now :-

$\sf m = \dfrac{h_i}{h _o }$

put :- m = -2 and ho = 7 cm

$\sf \implies - 2 = \dfrac{h_i}{7}$

$\sf \implies - 2 \times 7= {h_i}$

$\sf \implies - 14 \:cm= {h_i}$

Size of image is 14 cm. Negative sign shows that image is inverted and real.

Answer :

• screen be a placed at 54 cm from the pole of mirror ( in front of mirror) so that a sharp focused image can be obtained.

• Size of image = 14 cm

• Nature :- Real , inverted , magnified

Answer 2

Answer:

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Explanation: