Question

An object of height 7cm is kept at a distance of 25cm in front of a concave mirror focal length is 15cm at distance will we get a clear image ? what will be the size & nature of the imag​

Answer 1

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• An object of height 7cm is kept at a distance of 25cm in front of a concave mirror focal length is 15cm at distance will we get a clear image ? what will be the size & nature of the image?

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• Size of object h =7cm

• Size of the object h=?

• Focal length f=−15cm

• Object distance u=−25cm

• Image distance v=?

Using mirror formula

1/f=1/v+1/u

1/15=1/v +1/-25

1/v= -5+3/75

v = −37.5cm

Hence, position of image at 37.5cm infront of the mirror

Magnification(M)= h1/h2

= -v/u

=37.5/-25

= -1.5

h1=1.5×h2

= −1.5×7=−10.5cm

Hence,size of image is −10.5cm.

Nature of image is real,inverted and magnified.

Answer 2

Answer:

An object of height 7cm is kept at a distance of 25cm in front of a concave mirror focal length is 15cm at distance will we get a clear image ? what will be the size & nature of the image.

Explanation:

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Size of object h=7cm

Size of the object h =?

Focal length f =−15cm

Object distance u=−25cm

Image distance v=?

$using \: mirror \: formula \: \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{ - 15} = \frac{1}{v} + \frac{1}{ - 25}$

$\frac{1}{v} = \frac{ - 5 + 3}{ 75}$

$v = - 37.5 \: cm$

Hence, Position of image at 37.5cm infront of the mirror.

$Magnification \: (M) = \frac{h1}{h2} = \frac{ - v}{u} = \frac{37.5}{ - 25} = \: - 1.5$

$h1 = - 1.5 \times h2$

$h1 = - 1.5 \times \: 7 = - 10.5 \: cm$

Hence,size of image is −10.5cm.

Nature of image is real,inverted and magnified.