Question 365: Waves Optics

Question

In biprims experiment, the $$4^{	ext {th }}$$ dark band is formed opposite to one of the slits. The wavelength of light used is $$(\mathrm{d}=$$ distance between the slits, $$\mathrm{D}=$$ distance between scource and the screen)

Accepted Answer

In the biprism experiment, the formula for the position of dark bands in an interference pattern is given by the equation:
$$ 	ext{y}_	ext{dark} = (2n - 1) \frac{\lambda D}{2d} $$
where:
$$y_	ext{dark}$$: Position of the nth dark band
$$\lambda$$: Wavelength of light
$$D$$: Distance between the source and the screen
$$d$$: Distance between the slits
$$n$$: Order of the dark band
According to the problem, the 4th dark band is formed opposite to one of the slits. For constructive interference to form right at a slit, the position can be approximated as:
$$ y = \frac{d}{2} $$
Since we are considering the 4th dark band, $$n = 4$$. Substituting into the formula:
$$ y_{	ext{dark}}(4) = (2 	imes 4 - 1) \frac{\lambda D}{2d} $$
Therefore:
$$ \frac{d}{2} = (7) \frac{\lambda D}{2d} $$
Now, solve for $$\lambda$$:
$$ \lambda = \frac{d^2}{7D} $$
Thus, the wavelength of the light used in the experiment is:
Option B: $$ \frac{d^2}{7D} $$

Answer

$$\frac{\mathrm{d}^2}{14 \mathrm{D}}$$

Answer

$$\mathrm{\frac{d^2}{7 D}}$$

Answer

$$\frac{d^2}{9 \mathrm{D}}$$

Answer

$$\frac{d^2}{11 \mathrm{D}}$$

Waves Optics

In biprims experiment, the $4^{\text {th }} dark band is formed opposite to one of the slits. The wavelength of light used is (\mathrm{d}= distance between the slits, \mathrm{D}=$ distance between scource and the screen)

Choose the correct answer: