- (c) 9mm
Explanation:
Here, $\text{d} = 0.1 \text{mm},\lambda= 6000 \mathring{\text{A}}, \text{D}=0.5\text{m}$
For third dark band, $\text{d}\sin\theta=3\lambda\ ;\ \sin\theta=\frac{3\lambda}{\text{d}}=\frac{\text{y}}{\text{D}}$
$\text{y}=\frac{3\text{D}\lambda}{\text{d}}=\frac{3\times0.5\times6\times10^{-7}}{0.1\times10^{-3}}$
$=9\times10^{-3}\text{m}=9\text{mm}$
- (b) 10-2m
Explanation:
Given d = 0.2mm = 0.2 × 10-3m, D = 2m
$\lambda=5000\mathring{\text{A}}=5\times10^{-7}\text{m}$
The distance between the first minimum on other side of the central maximum
$\text{x}=\frac{2\lambda\text{D}}{\text{d}}=\frac{2\times5\times10^{-7}\times2}{0.2\times10^{-3}}$
$\Rightarrow\text{x}=10^{-2}\text{m}$
- (a) 6 × 10-3rad
Explanation:
Here, $\lambda=600\text{nm}=6\times10^{-7}\text{m}$
$\text{a}=0.2\text{mm}=2\times10^{-4}\text{m},\theta=?$
Angular width of central maxima,
$\theta=\frac{2\lambda}{\text{a}}=\frac{2\times6\times10^{-7}}{2\times10^{-4}}=6\times10^{-3}\text{rad}$
- (d) Diffraction bands become narrower and crowded together.
Explanation:
When red light is replaced by blue light $(\lambda_\text{B}<\lambda_\text{R})$ the diffraction pattern bands becomes narrow and crowded together.
- (b) Should be of the order of wavelength.
Explanation:
To observe diffraction, the size of the obstacle should be of the order of wavelength.