Ray Optics

JEE Main level, single correct, medium

157 of 295

Question 157 of 295

The position of the image formed by the combination of lenses is :

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Answer: (D) 30 cm30 \mathrm{~cm} (right of third lens)

For lens 1:f1=10,u=−30,v=1: \mathrm{f}_1=10, \mathrm{u}=-30, \mathrm{v}= ?
v=ufu+f=−30×10−30+10=15v=\frac{u f}{u+f}=\frac{-30 \times 10}{-30+10}=15
For lens 2: f1=−10,u=10,v=\mathrm{f}_1=-10, \mathrm{u}=10, \mathrm{v}= ?
v=ufu+f=10×−1010−10=∞v=\frac{u f}{u+f}=\frac{10 \times-10}{10-10}=\infty
For lens 3:f=30,u=−∞,v=3: \mathrm{f}=30, \mathrm{u}=-\infty, \mathrm{v}= ?
So v will be 30 .

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