JEE Main 12 Jan 2019 Morning Question 1
Question: What is the position and nature of image formed by lens combination shown in figure? ( $ f _1,f _2 $ are focal lengths) [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 70 cm from point B at right; real
B) $ \frac{20}{3}cm $ from point B at right, real
C) 40 cm from point B at right, real
D) 70 cm from point B at left, virtual
Show Answer
Answer:
Correct Answer: A
Solution:
-
For the first lens, $ \frac{1}{v _1}-\frac{1}{u _1}=\frac{1}{f _1} $
$ \Rightarrow $ $ \frac{1}{v _1}+\frac{1}{20}=\frac{1}{5}\Rightarrow v _1=\frac{20}{3}=6.67cm $
Now, for the second lens, $ u _2=6.67-2=\frac{14}{3}cm $
$ \frac{1}{v _2}-\frac{1}{u _2}=\frac{1}{f _2}\Rightarrow \frac{1}{v _2}=\frac{1}{-5}+\frac{3}{14} $ $ v _2=70cm $ right of second lens.
