Convex lens with two materials

JEE Advanced 2019 Paper 1, Question 10

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,

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  1. \bigg| \frac{\Delta f}{f} \bigg| < \bigg| \frac{\Delta n}{n} \bigg|
  2. 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 2^{\rm nd} decimal place).
  3. If \frac{\Delta n}{n} < 0 then \frac{\Delta f}{f} > 0
  4. The relation between \frac{\Delta f}{f} and \frac{\Delta n}{n} remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

Related articles:
Lenses I: The thin lens equation
Lenses II: Image formation


The Lens maker’s formula is

(1)   \begin{equation*}   \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \end{equation*}

where R_1 is the radius of curvature the object facing side (we keep the object on the left/front) and R_2 is the radius of the other side. By convention, R_i is positive when the center of curvature is in the back of the lens and negative otherwise. For instance, applying this convention to the lens in the figure below we can see that focal length of a convex lens is always positive. If the radii of the the two surfaces are equal in magnitude and n_1=n_2=n, (1) gives

(2)   \begin{equation*}   \frac{1}{f} = {\frac{2(n-1)}{R}} . \end{equation*}

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When n_1 \neq n_2 the lens can be thought of as two plano-convex lenses in series. The focal length of such an arrangement is

(3)   \begin{equation*}   \frac{1}{f_{\rm eff}} = \frac{1}{f_1} + \frac{1}{f_2}  \end{equation*}

where the focal length of each plano-convex lens can be obtained using \eqref{eq:q10LensMaker} by setting one of the radii to \infty, corresponding to the plane face. That is,

(4)   \begin{align*}   \frac{1}{f_1} &= (n_1-1) \left( \frac{1}{R}-\frac{1}{\infty} \right) = \frac{n-1}{R} \\   \frac{1}{f_2} &= (n_2-1) \left( \frac{1}{\infty}-\frac{1}{-R} \right) = \frac{n + \Delta n - 1}{R} . \end{align*}

Substituting these in (3)

(5)   \begin{equation*}   \frac{1}{1 + \frac{\Delta f}{f}} = {\frac{2(n-1)}{R}} + \frac{\Delta n}{R} \equiv \frac{1}{f} + \frac{\Delta n}{R} \end{equation*}

where we’ve set the effective focal length to f_{\rm eff} = f + \Delta f as given in the problem. Multiplying both sides of this equation by f we get

(6)   \begin{align*}   \frac{f}{f + \Delta f} = 1 + \Delta n \frac{f}{R} = 1 + \frac{\Delta n}{2(n-1)} . \end{align*}

For a small change in refractive index between n_1 and n_2, the change \Delta f \ll f. Therefore we can expand the expression on the left side using \frac{1}{1+x} \approx 1-x for x \ll 1,

(7)   \begin{equation*}   1 - \frac{\Delta f}{f} = 1 + \frac{\Delta n}{2(n-1)}   \implies   -\frac{\Delta f}{f} = \frac{\Delta n}{2(n-1)} . \end{equation*}

To relate \frac{\Delta n}{2(n-1)} to \frac{\Delta n}{n}, we note that

(8)   \begin{equation*}   2(n-1) - n = n-2 < 0 \end{equation*}

since n<2. That means n > 2(n-1), which implies \frac{\Delta n}{2(n-1)} > \frac{\Delta n}{n}. Thus,

(9)   \begin{equation*}   -\frac{\Delta f}{f} > \frac{\Delta n}{n}  \end{equation*}

The relation between the absolute values of these ratios is more subtle. To see this note that (7) relates the sign of \Delta f and \Delta n (recall that f and n-1 are positive). For \Delta n > 0 we have \Delta f < 0 and (9) implies

(10)   \begin{equation*}   \bigg| \frac{\Delta f}{f} \bigg| > \bigg| \frac{\Delta n}{n} \bigg| \end{equation*}

On the other hand, if \Delta n < 0 we have \Delta f > 0, and (9) tells us

(11)   \begin{equation*}   \bigg| \frac{\Delta f}{f} \bigg| < \bigg| \frac{\Delta n}{n} \bigg| \end{equation*}

So (A) is not always true, so we take that option to be incorrect. We also saw from this discussion that if \frac{\Delta n}{n} < 0 then \frac{\Delta f}{f} > 0, which means (C) is correct.

For n=1.5, \Delta n = 10^{-3} and f = 20 cm, (7) gives |\Delta f| = 0.02 cm. So option (B) is correct.

Finally, changing both convex surfaces to concave simply changes R \to -R in the equations above. While this changes the sign of f, it does not alter (7) which means the relation (9) between \frac{\Delta f}{f} and \frac{\Delta n}{n} remains unaffected. So (D) is correct.

Thus, the correct answers are (B), (C) and (D).


When the surfaces are concave \Delta f and \Delta n will have the same sign by (7), since -f > 0. Does this change the relationship between the absolute values \bigg| \frac{\Delta f}{f} \bigg| and \bigg| \frac{\Delta n}{n} \bigg|?

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