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Geometrical Optics question

2014 · Shift 0 · Q53
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Geometrical Optics question

2014 · Shift 0 · Q53

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A thin convex lens made from crown glass (μ=32)\left( {\mu = {3 \over 2}} \right)(μ=23​) has focal length fff. When it is measured in two different liquids having refractive indices 43{4 \over 3}34​ and 53,{5 \over 3},35​, it has the focal lengths f1{f_1}f1​ and f2{f_2}f2​ respectively. The correct relation between the focal lengths is :
  1. A
    f1=f2<f{f_1} = {f_2} \lt ff1​=f2​<f
  2. B
    f1>f{f_1} \gt ff1​>f and f2{f_2}f2​ becomes negative
  3. C
    f2>f{f_2} \gt ff2​>f and f1{f_1}f1​ becomes negative
  4. D
    f1 {f_1}\,f1​ and f2 {f_2}\,f2​ both become negative
View written solutionFree

Correct answer: B

  1. Use lens maker formula in a medium

For a thin lens immersed in a medium of refractive index μm\mu_mμm​,

1fm=(μℓμm−1)(1R1−1R2)\frac{1}{f_m}=\left(\frac{\mu_{\ell}}{\mu_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)fm​1​=(μm​μℓ​​−1)(R1​1​−R2​1​)

where μℓ\mu_{\ell}μℓ​ is the refractive index of the lens material.

Here, the lens is made of crown glass:

μℓ=32\mu_{\ell}=\frac{3}{2}μℓ​=23​

Since the lens is convex in air, its geometrical factor

(1R1−1R2)>0\left(\frac{1}{R_1}-\frac{1}{R_2}\right)>0(R1​1​−R2​1​)>0

So the sign of focal length depends on

(μℓμm−1)\left(\frac{\mu_{\ell}}{\mu_m}-1\right)(μm​μℓ​​−1)
  1. Focal length in air

In air, μm=1\mu_m=1μm​=1, so

1f=(32−1)(1R1−1R2)=12(1R1−1R2)\frac{1}{f}=\left(\frac{3}{2}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right) =\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(23​−1)(R1​1​−R2​1​)=21​(R1​1​−R2​1​)

Thus f>0f>0f>0.

  1. Lens in liquid of refractive index 43\frac{4}{3}34​

Let focal length be f1f_1f1​.

1f1=(3243−1)(1R1−1R2)\frac{1}{f_1}=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​1​=(34​23​​−1)(R1​1​−R2​1​)

Compute:

3243=32⋅34=98\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{3}{2}\cdot\frac{3}{4}=\frac{9}{8}34​23​​=23​⋅43​=89​

So,

1f1=(98−1)(1R1−1R2)=18(1R1−1R2)\frac{1}{f_1}=\left(\frac{9}{8}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right) =\frac{1}{8}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​1​=(89​−1)(R1​1​−R2​1​)=81​(R1​1​−R2​1​)

Compare with air:

1f=12(1R1−1R2)\frac{1}{f}=\frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=21​(R1​1​−R2​1​)

Hence,

1/f11/f=1/81/2=14⇒f1=4f\frac{1/f_1}{1/f}=\frac{1/8}{1/2}=\frac{1}{4} \Rightarrow f_1=4f1/f1/f1​​=1/21/8​=41​⇒f1​=4f

Therefore,

f1>ff_1>ff1​>f

and it is still positive.

  1. Lens in liquid of refractive index 53\frac{5}{3}35​

Let focal length be f2f_2f2​.

1f2=(3253−1)(1R1−1R2)\frac{1}{f_2}=\left(\frac{\frac{3}{2}}{\frac{5}{3}}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)f2​1​=(35​23​​−1)(R1​1​−R2​1​)

Compute:

3253=32⋅35=910\frac{\frac{3}{2}}{\frac{5}{3}}=\frac{3}{2}\cdot\frac{3}{5}=\frac{9}{10}35​23​​=23​⋅53​=109​

So,

\frac{1}{f_2}=\left(\frac{9}{10}-1\right) left(\frac{1}{R_1}-\frac{1}{R_2}\right) =-\frac{1}{10}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Thus,

f2<0f_2<0f2​<0

So the convex lens behaves like a diverging lens in this liquid.

  1. Check options
  • A: f1=f2<ff_1=f_2<ff1​=f2​<f ❌
  • B: f1>ff_1>ff1​>f and f2f_2f2​ becomes negative ✅
  • C: f2>ff_2>ff2​>f and f1f_1f1​ becomes negative ❌
  • D: f1f_1f1​ and f2f_2f2​ both become negative ❌
  1. Final answer

The correct relation is:

f1>f and f2 is negative\boxed{f_1>f \text{ and } f_2 \text{ is negative}}f1​>f and f2​ is negative​

So the correct option is B.

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