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A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
CBSE | Class 12 | NCERT | Physics | Ray Optics and Optical Instruments
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?

Answer :

According to the question, the object given is virtual and the image formed is real.

Object distance is u= +12 cm

(i) The focal length of the convex lens is f =20 cm

Image distance is denoted by v

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

$ \frac{1}{v}-\frac{1}{12}=\frac{1}{20}$

$ \frac{1}{v}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}$

$ Therefore\;v=\frac{60}{8}=7.5cm$

Therefore, the image formed will be 7.5cm away from the lens, to its right.

(ii) Focal length of the concave lens is f =-16 cm

Image distance is denoted by v

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

$ \frac{1}{v}=-\frac{1}{16}+\frac{1}{12}$

$ =\frac{-3+4}{48}=\frac{1}{48}$

∴ v=48cm.

Therefore, the image so formed is 48cm away from the lens towards its right.

i

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