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A divergent lens has a focal length of 20cm.at what distance should an object of height 4cm from the optical centre of the lens be placed so that its image is formed 10cm away from the lens.find the size of the image also

Posted by Baiju Joseph 7 years, 3 months ago

CBSE > Class 10 > Science

2 answers

Shruti Sharma 7 years, 1 month ago

DIVERGENT LENS: CONCAVE LENS-

f=-20cm

v=-10cm because a concave lens always forms a virtual image on the same side as the object.

h_o=4cm

1/f = 1/v - 1/u [ Lens Formula]

-1/20 = - 1/u - 1/10

-1/u = 1/10 - 1/20

-1/u = (2-1)/20

-1/u = 1/20

u= - 20 cm

m = v/u = h_i/h_o

1/2 = h_i/4

h_i = 2cm

0Thank You

Naveen Sharma 7 years, 3 months ago

Ans. Distance between the object and the lens (u )= -10cm
Focal length (f) = 20 cm
Distance between the image and the lens v = ?

Using Lens Formula

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

=> $\frac{1}{20} = \frac{1}{v} - \frac{1}{- 10}$

=> $\frac{1}{v} = \frac{1}{20} - \frac{1}{10}$

=> $\frac{1}{v} = \frac{1 - 2}{20}$

=> $\frac{1}{v} = \frac{- 1}{20}$

=> v = -20 cm

The image is formed on the same side as the object and since v is negative the image is erect and virtual and is formed at a distance 20 cm in front of the lens.

We know that

$m = \frac{h_{1}}{h_{0}} = \frac{v}{u}$

height of the object h_o = 4 cm

=> $h_{1} = \frac{v \times h_{0}}{u}$

=> $h_{1} = \frac{- 20 \times 4}{- 10}$

=> $h_{1} = 8$

height of image = 8 cm

1Thank You

ANSWER

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