Question

A monochromatic light is incident on a single slit of width 0.014 mm. The angular position of the second bright line observed is \(2.81^\circ\). Find the wavelength of the incident light. \([ \sin(2.81^\circ) = 0.049072 ]\)

• A. 4026
• B. 5890
• C. 6040
• D. 2748

Hint:

Single-slit bright fringe approximate: \(\lambda = \frac{a \sin \theta}{m}\). Convert slit width to meters, angle in radians or use sine.

Answer 1

The Correct Option is D

Step 1: Write the bright fringe condition for a single slit.
\[ a\sin\theta = m\lambda \]
with slit width $a = 0.014\ \text{mm} = 1.4\times10^{-5}\ \text{m}$, order $m=2$, and $\sin(2.81^\circ)=0.049072$.
Step 2: Isolate the wavelength.
\[ \lambda = \frac{a\sin\theta}{m} = \frac{(1.4\times10^{-5})(0.049072)}{2} \]
Step 3: Multiply, then divide.
\[ (1.4\times10^{-5})(0.049072) = 6.870\times10^{-7}\ \text{m}, \qquad \lambda = \frac{6.870\times10^{-7}}{2} = 3.435\times10^{-7}\ \text{m} \]
Step 4: Convert metres to angstroms ($1\ \text{m} = 10^{10}\ \text{\AA}$).
\[ \lambda \approx 2748\ \text{\AA} \]
\[ \boxed{2748\ \text{\AA}} \]