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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2014 · Paper 2 · Physics · Ray Optics

Four combinations of two thin lenses are given in List I. The...

Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.
 

 List I List II
A.
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the
P.P .2r2r
B.
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the
Q.Q .r2\dfrac{r}{2}
C.
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the
R.R .−r- r
D.
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the
S.S .rr

Options

  1. Aa-r;b-p;c-s;d-q;
  2. Ba-s;b-r;c-q;d-p;
  3. Ca-p;b-q;c-s;d-r;
  4. Da-q;b-s;c-r;d-p;

Correct answer

D. a-q;b-s;c-r;d-p;

Step-by-step solution

1f=1f1+1f2\dfrac{1}{f} = \dfrac{1}{{f}_{1}} + \dfrac{1}{{f}_{2}}

Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the
Use 1feq=1f1+1f2\dfrac{1}{{f}_{eq}} = \dfrac{1}{{f}_{1}} + \dfrac{1}{{f}_{2}}
\left(P\right)\dfrac{1}{{f}_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{2}{R}; {f}_{eq} = \dfrac{R}{2}<br>(Q)1feq=12R+12R=1R;feq=R<br >\left(Q\right)\dfrac{1}{{f}_{eq}} = \dfrac{1}{2R} + \dfrac{1}{2R} = \dfrac{1}{R}; {f}_{eq} = R
(R)1feq=−12R−12R=−1R;feq=−R\dfrac{\left(R\right)1}{{f}_{eq}} = - \dfrac{1}{2R} - \dfrac{1}{2R} = - \dfrac{1}{R}; {f}_{eq} = - R
(S)1feq=1R−12R=12R;feq=2R\dfrac{\left(S\right)1}{{f}_{eq}} = \dfrac{1}{R} - \dfrac{1}{2R} = \dfrac{1}{2R}; {f}_{eq} = 2R
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