JEE Advanced 2019 · Paper 1 · Physics Q10

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PhysicsMultiple correct+4 / -1Question 10 of 54

A thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$ , as shown in figure. The radius of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1 = n_2 = n$ . The focal length is $f + \Delta f$ when $n_1 = n$ and $n_2 = n + \Delta n$ . Assuming $\Delta n \ll (n-1)$ and $1 < n < 2$ , the correct statement(s) is/are,

$A thin biconvex lens split by a vertical dashed line along its axis into a left half of refractive index n₁ and a right half of refractive index n₂.$

\left|\dfrac{\Delta f}{f}\right| < \left|\dfrac{\Delta n}{n}\right|

For

n = 1.5

\Delta n = 10^{-3}

and

f = 20

cm, the value of

|\Delta f|

will be 0.02 cm (round off to 2nd decimal place).

\dfrac{\Delta n}{n} < 0

then

\dfrac{\Delta f}{f} > 0

The relation between

\dfrac{\Delta f}{f}

and

\dfrac{\Delta n}{n}

remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

Answer and worked solution

Official answer: B, C, D

Power of the lens. Each half is a plano-convex lens of radius $R$ , and the two are in contact:
$\frac1f=\frac{(n_1-1)+(n_2-1)}R.$
With $n_1=n_2=n$ , $\frac1f=\frac{2(n-1)}R$ .

First-order change. Changing $n_2$ by $\Delta n$ :
$-\frac{\Delta f}{f^2}=\frac{\Delta n}R\implies\frac{\Delta f}f=-\frac{\Delta n}{2(n-1)}.$

(A) For $1<n<2$ , $2(n-1)<n$ , so $\left|\frac{\Delta f}f\right|>\left|\frac{\Delta n}n\right|$ . False.
(B) $|\Delta f|=\dfrac{20\times10^{-3}}{2(0.5)}=0.02$ cm. True.
(C) The signs are opposite, so $\Delta n<0$ gives $\Delta f>0$ relative to $f$ . True.
(D) Concave surfaces flip the sign of $R$ and of $f$ , but $\frac{\Delta f}f=-\frac{\Delta n}{2(n-1)}$ is unchanged. True.

(B), (C), (D)

Topics:Thin lenses and lens maker's formula Combination of lenses and mirrors

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JEE Advanced 2019 · Paper 1, Physics question 10