If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$
For $n = 1.5,\; \Delta n = 10^{-3}$ and $f = 20\,\mathrm{cm}$ ,the value of $|\Delta f|$ will be 0.02 cm (rounded off to 2nd decimal place).
The relation between $\;\; \frac{\Delta f}{f}$ and $\;\; \frac{\Delta n}{n}$ remains unchanged if both the convex surfaces arereplaced by concave surface of the same radius of curvature.
$\left|\;\; \frac{\Delta f}{f}\right| < \left|\;\; \frac{\Delta n}{n}\right|$

Solution:

f = \;\; \frac{R}{2(n-1)}

...........(1)

In 2nd case

\;\; \frac{1}{f+df} = \;\; \frac{n-1}{R} + \;\; \frac{(n+dn)-1}{R}

\;\; \frac{1}{f+df} = \;\; \frac{1}{f} + \;\; \frac{dn}{R}

Clearly,if

dn < 0

,R.H.S value is less than

\frac{1}{f}

\therefore df > 0

i.e (1) is correct

Option(2)

f = \;\; \frac{R}{2(1.5-1)} \Rightarrow R = 20\,\mathrm{cm}

\;\; \frac{1}{f+df} = \;\; \frac{1}{f} + \;\; \frac{dn}{R}

\;\; \frac{1}{20+df} = \;\; \frac{1}{20} + \;\; \frac{dn}{20}

\;\; \frac{1}{20\left(1 + \;\; \frac{df}{20}\right)} = \;\; \frac{1}{20}(1+dn)

1 - \;\; \frac{df}{20} = 1 + dn \Rightarrow df = -20\,dn

= -20 \times 10^{-3}

\therefore |df| = 0.02

(2) is correct

Option(3): If convex surfaces are replaced with concave surfaces, sign of R will change

\;\; \frac{1}{f\left(1 + \;\; \frac{df}{f}\right)} = \;\; \frac{1}{f} + \;\; \frac{dn}{R} = \;\; \frac{1}{f} + \;\; \frac{dn}{2f(n-1)} \begin{pmatrix} f = \;\; \frac{R}{2(n-1)} \\ \;\; \frac{1}{R} = \;\; \frac{1}{2f(n-1)} \end{pmatrix}

\;\; \frac{1 - \;\; \frac{df}{f}}{f} = \;\; \frac{1}{f}\left[1 + \;\; \frac{dn}{2n\left(1 - \;\; \frac{1}{n}\right)}\right]

- \;\; \frac{df}{f} = \left(\;\; \frac{dn}{n}\right) \;\; \frac{1}{2\left(1 - \;\; \frac{1}{n}\right)}

As this is independent of R, (3) is correct

Option (4):

\left|\;\; \frac{df}{f}\right| = \left|\;\; \frac{dn}{n}\right| \;\; \frac{1}{2\left(1 - \;\; \frac{1}{n}\right)}

For the range

1 < n < 2,\; 2\left(1 - \;\; \frac{1}{n}\right)

is always greater than 1 i.e,

\left|\;\; \frac{df}{f}\right| > \left|\;\; \frac{dn}{n}\right|

∴ (4) is wrong

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