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QJEEAdv16Ph210d: JEE Advanced 2016 Physics - Class 12 JEEAdv16Ph210d Solution: A plano-convex lens is made of material of refractive index n.…

A plano-convex lens is made of material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away form the lens. Which of the following statement(s) is(are) true?

Multiple Choice Options

  • A The refractive index of the lens is 2.5
  • B The radius of curvature of the convex surface is 45 cm
  • C The faint image is erect and real
  • D The focal length of the lens is 20 cm

Step-by-Step Verified Solution

For refraction through plano - Convex lens,

m=−2=vum=-2=\frac{v}{u}

⇒v=−2u As, u=−30 cm⇒v=60 cm\begin{array}{ll}\Rightarrow v=-2 u \\\\\text { As, } u=-30 \mathrm{~cm} \\\\\Rightarrow v=60 \mathrm{~cm}\end{array}

Using lens formula, 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

160−1−30amp;=1f⇒f=20 cm\begin{aligned}\frac{1}{60}-\frac{1}{-30} & =\frac{1}{f} \\\\\Rightarrow f =20 \mathrm{~cm}\end{aligned}

JEE Advanced 2016 Paper 1 Offline Physics - Geometrical Optics Question 44 English Explanation 1

By lens maker's formula,

1famp;=(n−1)(1R1−1R2)⇒120amp;=(n−1)(1R+1∞)=n−1R⇒namp;=1+(R20)........(i)\begin{aligned}\frac{1}{f} & =(n-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\\\\Rightarrow \frac{1}{20} & =(n-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\infty}\right)=\frac{n-1}{\mathrm{R}} \\\\\Rightarrow n & =1+\left(\frac{\mathrm{R}}{20}\right) ........(i)\end{aligned}

for poor reflection from convex surface,

amp;u=−30 cm,amp;v=10 cm,f=R/2\begin{aligned}& u=-30 \mathrm{~cm}, \\\\& v=10 \mathrm{~cm}, f=\mathrm{R} / 2\end{aligned}

JEE Advanced 2016 Paper 1 Offline Physics - Geometrical Optics Question 44 English Explanation 2

Using mirror formula,

1v+1uamp;=1f=2R;⇒110+1−30amp;=2R\begin{aligned}\frac{1}{v}+\frac{1}{u} & =\frac{1}{f}=\frac{2}{\mathrm{R}} ; \\\\\Rightarrow \frac{1}{10}+\frac{1}{-30} & =\frac{2}{\mathrm{R}}\end{aligned}

⇒2R=230∴R=30 cm\begin{array}{llrl}\Rightarrow \frac{2}{\mathrm{R}} =\frac{2}{30} \\\\\therefore \mathrm{R} =30 \mathrm{~cm}\end{array}

Substitute R=30 cmR=30 \mathrm{~cm} in equation (i) to get n=n= 2.5 for f=20 cmf=20 \mathrm{~cm} (inverted image) and n=1.5n=1.5 for f=60 cmf=60 \mathrm{~cm} (erect image).

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