Physics, asked by jayaramchowdhary, 11 months ago

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​

Answers

Answered by amanreddypeddireddy
2

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

Answered by swethassynergy
0

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.

#SPJ2

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