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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2012 · Paper 1 · Physics · Ray Optics

A bi-convex lens is formed with two thin plano-convex lenses as...

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index nn of the first lens is 1.51.5 and that of the second lens is 1.21.2. Both the curved surface are of the same radius of curvature R=14R=14 cm\mathrm{cm}. For this bi-convex lens, for an object distance of 40 cm\mathrm{cm}, the image distance will be A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first

Options

  1. A−280.0 cm-280.0 \mathrm{~cm}
  2. B40.0 cm40.0 \mathrm{~cm}
  3. C21.5 cm21.5 \mathrm{~cm}
  4. D13.3 cm13.3 \mathrm{~cm}

Correct answer

B. 40.0 cm40.0 \mathrm{~cm}

Step-by-step solution

The focal length (f1)\left(f_{1}\right) of the plano-cöneex Tens with n=1.5n=1.5 using lens-maker formula 1f1=(n1−1)[1R1−1R2]=(1.5−1)[114−1∞]=128\frac{1}{f_{1}}=\left(n_{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.5-1)\left[\frac{1}{14}-\frac{1}{\infty}\right]=\frac{1}{28} The focal length (f2)\left(f_{2}\right) of the plano-convex lens with n=1.21f2=(n2−1)[1R1−1R2]=(1.2−1)[1∞−1−14]=170n=1.2 \frac{1}{f_{2}}=\left(n_{2}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.2-1)\left[\frac{1}{\infty}-\frac{1}{-14}\right]=\frac{1}{70} Focal length F of the combination 1F=1f1+1f2=120\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{20} Now, applying lens formula for the combination of lens 1V−1U=1F⇒1V−1−40=120[ Given μ=40cm]∴V=40cm\begin{array}{ll} & \frac{1}{V}-\frac{1}{U}=\frac{1}{F} \Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20} \quad[\text{ Given } \mu=40 \mathrm{ cm}]\\ \therefore \quad & V=40 \mathrm{ cm} \end{array}
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