Question 360: Wave Optics

Question

In the interference experiment using a biprism, the distance of the slits from the screen is increased by $$25 \%$$ and the separation between the slits is halved. If '$$W$$' represents the original fringewidth, the new fringewidth is

Accepted Answer

To determine how the fringe width changes when the distance between the slits and the screen is increased by $$25 \%$$ and the separation between the slits is halved, we need to understand the formula for fringe width in a double-slit interference pattern:
The fringe width $$W$$ is given by:
$$ W = \frac{\lambda D}{d} $$
where:
$$\lambda$$ is the wavelength of the light
$$D$$ is the distance between the slits and the screen
$$d$$ is the separation between the slits
Originally, the fringe width is:
$$ W = \frac{\lambda D}{d} $$
Now let's consider the new conditions:
The distance between the slits and the screen is increased by $$25 \%$$, so the new distance $$D'$$ is:
$$ D' = D + 0.25D = 1.25D $$
The separation between the slits is halved, so the new separation $$d'$$ is:
$$ d' = \frac{d}{2} $$
The new fringe width $$W'$$ is then:
$$ W' = \frac{\lambda D'}{d'} $$
Substituting the new values:
$$ W' = \frac{\lambda \cdot 1.25D}{\frac{d}{2}} $$
This simplifies to:
$$ W' = \frac{1.25 \cdot 2 \cdot \lambda D}{d} $$
$$ W' = 2.5 \cdot \frac{\lambda D}{d} $$
Since $$\frac{\lambda D}{d} = W$$, we have:
$$ W' = 2.5 W $$
Therefore, the new fringe width is $$2.5W$$, corresponding to Option B:
Option B: 2.5 W

Answer

2 W

Answer

2.5 W

Answer

4 W

Answer

1.5 W

Wave Optics

In the interference experiment using a biprism, the distance of the slits from the screen is increased by $25 \% and the separation between the slits is halved. If 'W$' represents the original fringewidth, the new fringewidth is

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