Explanation

(d) Let $n$th minima of 400 nm coincides with mth minima of 560 nm, then
$(2n - 1)\left( \frac{400}{2} \right) = (2m - 1)\left( \frac{560}{2} \right)$
Or $\frac{2n - 1}{2m - 1} = \frac{7}{2} = \frac{14}{10} = \ldots$
$ie$. 4$^{th}$minima of 400 nm coincides with 3$^{rd}$ minima of 560 nm.
Location of this minima is,
$Y_{1} = \frac{(2 \times 4 - 1)(1000)\left( 400 \times 10^{- 6} \right)}{2 \times 0.4} = 14\ mm$
Next 11$^{th}$ minima of 400 nm will coincide with 8$^{th}$ minima of 560 nm. Location of this minima is ,
$Y_{2} = \frac{(2 \times 11 - 1)\ (1000)\ \left( 400 \times 10^{- 6} \right)}{2 \times 0.1} = 42\ mm$
$\therefore\ $Required distance$= Y_{2} - Y_{1} = 28$ mm