JEE Advanced 2025 — Paper 1 — Question 10

If we can consider$\frac{\Delta \mathrm{b}}{\mathrm{b}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \lambda}{\lambda}+\frac{\Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}$

$\frac{\Delta \mathrm{b}}{\mathrm{b}}=0+0+\frac{1 \mathrm{~cm}}{1 \mathrm{~m}}+\frac{1 \mathrm{~mm}}{5 \mathrm{~mm}}=0.21$

$\mathrm{b}=\frac{\mathrm{m} \lambda \mathrm{D}}{\mathrm{d}}=\frac{3 \times 600 \times 10^{-3} \times 1}{5 \times 10^{-3}} \mu \mathrm{~m}=360 \mu \mathrm{~m}$ $\Rightarrow \Delta \mathrm{b}=360 \times 0.21 \mu \mathrm{~m}=75.6 \mu \mathrm{~m}$

However, error in d is too large ( $20 \%$ ) for the

solution- 1 to be correct.

Hence, we propose

Solution-2

$\mathrm{b}=\frac{\mathrm{m} \lambda \mathrm{D}}{\mathrm{d}}=360 \mu \mathrm{~m}$

$\mathrm{b}_{\text {max }}=\frac{3 \times 600 \times 10^{-3} \times 1.01}{4 \times 10^{-3}} \mu \mathrm{~m}=454.5 \mu \mathrm{~m}$

$\mathrm{b}_{\text {min }}=\frac{3 \times 600 \times 10^{-3} \times 0.99}{6 \times 10^{-3}} \mu \mathrm{~m}=297 \mu \mathrm{~m}$

Maximum value of $b$ gives error, $\Delta b_{1}=94.5 \mu \mathrm{~m}$

Minimum value of b gives error, $\Delta \mathrm{b}_{2}=63 \mu \mathrm{~m}$

$\therefore$ We always report the largest error, hence correct answer should be $94.5 \mu \mathrm{~m}$