Physics 3 Marks Question

$\beta=\lambda\frac{\text{D}}{\text{d}}$

$\Rightarrow \beta=\frac{\lambda}{\text{d}}\text{D}$

The graph between β and D is shown alongside

The slope of graph $=\frac{\lambda}{\text{d}}$

Knowing d, the wavelength of light used can be calculated to be

$\lambda= \text{Slope of graph}\times \text{d.}$ 

1. The fringe width $\beta= \frac{\text{D}\lambda}{\text{d}}.$
2. When the width of the slit is doubled; the intensity of interfering waves becomes four times, intensity of maxima becomes 16 times i.e., fringes become brighter.
3. When separation between the slits is increased the fringe width decreases, i.e., fringes come closer.
4. $\beta \alpha\text {D }$ when screen is moved away from the plane of the slits, the fringe width increases, i.e., fringes become farther away.

1. Since, there is a phase difference of $\pi$ between direct light and reflecting light, the intensity just above the mirror will be zero.
2. Here, 2d = equivalent slit separation

D = Distance between slit and screen.

We know for bright fringe, $\Delta\text{x}=\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda$

But as there is a phase reversal of $\frac{\lambda}{2}.$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}+\frac{\lambda}{2}=\text{n}\lambda$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda-\frac{\lambda}{2}\Rightarrow\text{y}=\frac{\lambda\text{D}}{4\text{d}}$

So, $\frac{1472}{1}=\frac{3\times10^8}{\text{v}_{400}}\Rightarrow\text{v}_{400}=2.04\times10^8\text{m/sec}.$

[because, for air, $\mu=1\ \text{and v}=3\times10^8\text{m/s}]$

Again, $\frac{1452}{1}=\frac{3\times10^8}{\text{v}_{760}}\Rightarrow\text{v}_{760}=2.07\times10^8\text{m/sec}.$