Question

an object of 5 cm length is placed along the axis of a concave mirror of focal length 15 CM such that its near and is at 20 cm from the pole find the length of the image and magnification​

Answer 1

Explanation:

ho=5cm

f= -15cm

u= -20cm

1/v + 1/u = 1/f

1/v = 1/-15 -1/20

= v = 8.57 cm

m= -v/u

m = 0.4285

m = hi/ho

hi = 2.1425

Answer 2

The length of the image and magnification​ are 37.50 cm and 7.5 respectively.

Explanation:

Given:

An object of 5 cm length is placed along the axis of a concave mirror of focal length 15 CM.

Its near and is at 20 cm from the pole.

To Find:

The length of the image and magnification​.

Formula Used:

Mirror equation for concave mirror

$\frac{1}{u} +\frac{1}{v} =\frac{1}{f}$   -----------  equation no.01.

Where,

u= Object distance

v= Image distance

f= Focal length of concave mirror.

Magnification equation

$M=\frac{I}{O}$  ------------- equation no.02.

Where,

M= Magnification

I= Image length

O=Object length

Solution:

As given,an object of 5 cm length is placed along the axis of a concave mirror of focal length 15 CM.

I= 5 cm  ,  f=-15 cm  and u=-20 cm.

For end of the rod  image distance $u_{1}$=-20cm.

Applying the formula no.01.

$\frac{1}{(-20)} +\frac{1}{v_{1} } =\frac{1}{(-15)}$

$-\frac{1}{20} +\frac{1}{v_{1} } =-\frac{1}{15}$

$\frac{1}{v_{1} } =\frac{1}{20}-\frac{1}{15}$  

$\frac{1}{v_{1} } =\frac{3-4}{60}$

    $=-\frac{1}{60}$  

$v_{1} =-60cm$

For $u_{2}$= -25 cm

$\frac{1}{(-25)} +\frac{1}{v_{2} } =\frac{1}{(-15)}$

$-\frac{1}{25} +\frac{1}{v_{2} } =-\frac{1}{15}$

$\frac{1}{v_{2} } =\frac{1}{25}-\frac{1}{15}$

$\frac{1}{v_{2} } =\frac{3-5}{75}$

$v_{2} =-37.5cm$

Hence, Image length =$v_{2} -v_{1} = -37.50-(-60)$

                                                  $=60-37.50= 22.50cm$

Applying the formula no.02.

$M=\frac{37.50}{5} =7.5$

Thus,the length of the image and magnification​ are 37.50 cm and 7.5 respectively.

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