NVAMediumJEE 2026Refraction & Lenses

JEE Physics 2026 Question with Solution

A convex lens of refractive index 1.51.5 and focal length f=18cmf=18 \, \text{cm} is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α×f\alpha \times f. Find the value of α\alpha. (Given: refractive index of water =43=\dfrac{4}{3})

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: A convex lens has refractive index μlens=1.5\mu_{\text{lens}}=1.5 and focal length in air fair=18cmf_{\text{air}}=18 \, \text{cm}. The refractive index of water is μwater=43\mu_{\text{water}}=\dfrac{4}{3}.

Find: The value of α\alpha if the difference in focal lengths is αf\alpha f.

Concept: For a thin lens immersed in a medium, use the refractive index of the lens relative to the medium in the lens maker formula.

1fm=(μlensμmedium1)(1R11R2)\frac{1}{f_m}=\left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

In air,

1fair=(μ1)(1R11R2)\frac{1}{f_{\text{air}}}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Using fair=18cmf_{\text{air}}=18 \, \text{cm} and μ=1.5\mu=1.5,

(1R11R2)=118(1.51)=19\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{18(1.5-1)}=\frac{1}{9}

Now for water, the relative refractive index is

μrel=μlensμwater=1.54/3=98\mu_{\text{rel}}=\frac{\mu_{\text{lens}}}{\mu_{\text{water}}}=\frac{1.5}{4/3}=\frac{9}{8}

So,

1fwater=(981)19=1819=172\frac{1}{f_{\text{water}}}=\left(\frac{9}{8}-1\right)\frac{1}{9}=\frac{1}{8}\cdot\frac{1}{9}=\frac{1}{72}

Hence,

fwater=72cmf_{\text{water}}=72 \, \text{cm}

The difference in focal lengths is

Δf=fwaterfair=7218=54cm\Delta f=f_{\text{water}}-f_{\text{air}}=72-18=54 \, \text{cm}

Given that

Δf=αf\Delta f=\alpha f

therefore,

α=5418=3\alpha=\frac{54}{18}=3

Therefore, the value of α\alpha is 33.

Using relative refractive index explicitly

Given: The lens has refractive index 1.51.5, focal length in air 18cm18 \, \text{cm}, and water has refractive index 43\dfrac{4}{3}.

Find: The value multiplying ff in the focal length difference.

The key idea is that when the lens is placed in water, its power reduces because the contrast between lens material and surrounding medium decreases.

In air, lens power is proportional to

(1.51)=0.5(1.5-1)=0.5

In water, the effective refractive index factor becomes

1.54/31=981=18\frac{1.5}{4/3}-1=\frac{9}{8}-1=\frac{1}{8}

Thus the power in water is

1/81/2=14\frac{1/8}{1/2}=\frac{1}{4}

Since focal length is inversely proportional to power,

fwater=4fair=4×18=72cmf_{\text{water}}=4f_{\text{air}}=4\times 18=72 \, \text{cm}

Hence the change in focal length is

7218=54cm72-18=54 \, \text{cm}

Now,

54=α×1854=\alpha \times 18

so

α=3\alpha=3

Therefore, the correct numerical answer is 33.

Common mistakes

  • Using the refractive index of the lens directly in water is incorrect because the lens maker formula in a medium uses the relative refractive index μlensμmedium\dfrac{\mu_{\text{lens}}}{\mu_{\text{medium}}}. Always replace μ\mu by the relative value first.

  • Subtracting focal lengths in the wrong order can cause sign confusion. Here the question asks for the difference in focal lengths, and since the focal length in water is larger, use fwaterfairf_{\text{water}}-f_{\text{air}} to get the positive change.

  • Assuming focal length remains unchanged in water is wrong because the surrounding medium affects lens power. A smaller refractive index contrast means smaller power and hence a larger focal length.

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