GATE 2022 IN Q.51: solution and answer

Official answer: 0.75 to 0.85Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Light passing through a uniformly absorbing sample follows the Beer–Lambert law: $I(x)=I_0{e}^{-\alpha x}$, where $\alpha$ is the absorption coefficient and $I_0$ is the unknown incident intensity.

2. For the 1 cm sample: $I_1=I_0{e}^{-\alpha (1\text{cm})}=0.414{\text{mW/cm}}^2$.

3. For the 2 cm sample: $I_2=I_0{e}^{-\alpha (2\text{cm})}=0.186{\text{mW/cm}}^2$.

4. Divide the first equation by the second so that $I_0$ cancels: $$\frac{I_1}{I_2}=e^{\alpha (x_2-x_1)}=e^{\alpha (1\text{cm})}$$

5. Take the natural logarithm: $$\alpha =\frac{1}{x_2-x_1}\ln \frac{I_1}{I_2}=\frac{1}{1\text{cm}}\ln \frac{0.414}{0.186}$$

6. Evaluate the ratio: $\frac{0.414}{0.186}\approx 2.2258$, so $\ln 2.2258\approx 0.8001$.

7. Therefore $\alpha \approx 0.80{\text{cm}}^{-1}$. The wavelength of 532 nm is not needed in the calculation.