EXAMPLE PROBLEMS FROM REFLECTION OF LIGHT

 

EXAMPLE PROBLEMS:

QUESTIONS:

1)    An object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm. Then find image distance and nature of the image.

ANS: 

Mirror type: concave

Given data,  u=-20 cm( “-“ is because of sign convention, object distance should be measured from pole to object which is in opposite

Direction to the direction of incident ray)

   f=-10 cm ( applying sign convention same as above)

mirror formula  =>   1/f=1/u+1/v

ð 1/-10=(1/-20)+(1/v)

ð 1/v=(1/20)+(1/-10)

ð 1/v=(1/20)-(1/10)

ð 1/v=(1-2)/20 (on taking L.C.M)

ð 1/v=-1/20

ð V=-20 cm

Nature:         

Here “-“ sign indicates the image formed is real and inverted.

The image is formed on the object itself as the object distance and image distance are equal. The image is formed at C. (refer case 3 of concave mirror)

 

2)    An object is kept at a distance of 10 cm in front of a convex mirror of focal length 5 cm. Find the position and nature of the image.

ANS:

Mirror type: convex:

Given data, u=-10 cm

f=+5 cm ( focus will be inside of the mirror, so focal length is measured in the direction of incident ray from pole to focus ).

1/f=1/u+1/v

ð 1/v=(1/f)-(1/u)

ð 1/v=(1/5)-(1/-10)

ð 1/v=1/5+1/10

ð 1/v=2+1/10  (L.C.M)

ð 1/v=3/10

ð V=10/3 => v=3.33 cm (approximately)

Nature:

Here we got answer in “+” sign, therefore image formed is virtual and erect. Image distance is 3.33 cm so, image formed must be between P and F.

 

3)    If an object is at a distance of 4 cm in front of a concave mirror an image is formed at a distance of 8 cm inside the mirror. Then find its magnification and nature of the image.

ANS:

Mirror type: concave

Given u=-4 cm, v=+8 cm ( image is virtual and erect as it formed inside the mirror)

Magnification, m=-v/u

ð m=-8/-4

ð m=+2

here “+” indicates image is virtual and erect.

And as 2>1 image formed is of enlarged size ( refer magnification topic for clarity)

 

4)    If the focal length of a concave mirror is 5 cm, at what distance an object should be placed to get an inverted image of same size on the object itself.

ANS:

Mirror type: concave

Given  f=-5cm,

Also given object distance is equal to image distance

Let u=v=-x

1/f=(1/u)+(1/v)

ð 1/-5=(1/-x)+(1/-x)

ð 1/-5=2(1/-x)

ð  1/-5=2/-x

ð 1/5=2/x

ð X=10 cm

v=-x 

                 v=-10 cm.

5)    An object is placed at a distance of 8 cm in front a convex mirror and if a virtual image is formed at a distance of 6 cm inside the mirror, then find the focal length of the mirror and magnification of the image.

            ANS:

                 Mirror type: convex

            Given  u=-8 cm, v=+6 cm

            1/f=1/u+1/v

ð 1/f=(1/-8)+(1/6)

ð 1/f=(-6+8)/48  [L.C.M]

1/f=2/48

f=24 cm

magnification  m=-v/u

m=-6/-8

m=3/4

0r m=0.75

The magnification obtained is “+”, so image is virtual and erect

m<1 (0.75<1), so image is diminished.

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