MCQ
In a single slit diffraction experiment first minima of $\lambda_1=6000 Å$ coincides with first maxima for wavelength $\lambda_2$, then $\lambda_2$ is :
- ✓4000 Å
- B5000 Å
- C4800 Å
- D5500 Å
✓
Answer
Correct option: A.
4000 Å
(A) 4000 Å,
For first minima of $\lambda_1, a \sin \theta=1 . \lambda_1$
For first maxima of $\lambda_2$, a $\sin \theta=3 / 2 \lambda_2$
$\therefore \quad \lambda_1=\frac{3}{2} \lambda_2$
or $\quad \lambda_2=\frac{2}{3} \times \lambda_1=\frac{2}{3} \times 6000 \Rightarrow 4000 Å$
For first minima of $\lambda_1, a \sin \theta=1 . \lambda_1$
For first maxima of $\lambda_2$, a $\sin \theta=3 / 2 \lambda_2$
$\therefore \quad \lambda_1=\frac{3}{2} \lambda_2$
or $\quad \lambda_2=\frac{2}{3} \times \lambda_1=\frac{2}{3} \times 6000 \Rightarrow 4000 Å$
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