Question

24. An object of height 4 cm is placed perpendicular to the principal axis of a concave
lens of local length 10 cm. If object is kept at a distance 20 cm from optical centre,
then find the size of image.
1​

Answer 1

Answer:

Size of the image = 1.33 cm

Explanation:

Given:

• Height of the object = 4 cm
• Focal length of concave lens = -10 cm
• Object distance = -20 cm

To Find:

• Height of the image

Solution:

By lens formula we know that,

$\sf \dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u}$

where f is the focal length, v is the image distance and u is the object distance.

Substituting the data we get,

$\sf \dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-20}$

$\sf \dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{-20}$

$\sf \dfrac{1}{v} =\dfrac{-3}{20}$

$\sf v=\dfrac{-20}{3} \:cm$

Now we know that,

$\sf m=\dfrac{v}{u} =\dfrac{h'}{h}$

where m is the magnification, h' is the height of the image and h is the height of the object.

Substitute the data,

$\sf \dfrac{-20}{3} \div-20=\dfrac{h'}{4}$

$\sf \dfrac{h'}{4} =\dfrac{1}{3}$

$\sf 3\:h'=4$

$\sf h'=\dfrac{4}{3}\:cm=1.33\:cm$

Hence the size/height of the image is 1.33 cm.

Answer 2

Answer:

Given :-

An object of height 4 cm is placed perpendicular to the principal axis of a concave

lens of local length 10 cm. If object is kept at a distance 20 cm from optical centre

To Find :-

Size of image

Solution :-

We know that

1/v - 1/u = 1/f

1/-10 + 1/v = 1/-20

-1/10 + 1/v = -1/20

1/v = -1/10 + (-1)/20

1/v = -1/10 - 1/20

1/v = -2 - 1/20

1/v = -3/20

v = -20/3 = -6.66

Now

m = -v/u = hi/ho

-6.66/-20 = hi/4

6.66/20 = hi/4

20 × hi = 6.66 × 4

20hi = 26.64

hi = 1.33