wo slits separated by a distance of 1 mm, are illuminated with red light of wavelength 6.5 x10-7 m. The interference fringes are observed on a screen placed 1 m from the slits. Find the distance (in mm) between the third dark fringe and the fifth bright fringe on the same side of the central maxima.Correct answer is '1.625'. Can you explain this answer?

Question

Answer

To find the distance between the third dark fringe and the fifth bright fringe, we can use the formula for the fringe spacing in Young's double-slit experiment:

$$d = \frac{{\lambda L}}{{D}}$$

where:
- $d$ is the fringe spacing (distance between consecutive fringes)
- $\lambda$ is the wavelength of light
- $L$ is the distance between the slits and the screen
- $D$ is the distance between the slits

Let's calculate the fringe spacing first:

$$\begin{align*}
d &= \frac{{6.5 \times 10^{-7} \, \text{m} \times 1 \, \text{m}}}{{1 \, \text{mm}}} \
&= 6.5 \times 10^{-4} \, \text{m} \, \text{or} \, 0.65 \, \text{mm}
\end{align*}$$

This means that the distance between consecutive fringes is 0.65 mm.

Now we need to find the distance between the third dark fringe and the fifth bright fringe on the same side of the central maximum.

$$\begin{align*}
\text{Distance between the central maximum and the third dark fringe} &= 2 \times (\text{distance between consecutive fringes}) \times \text{number of fringes} \
&= 2 \times 0.65 \, \text{mm} \times 3 \
&= 3.9 \, \text{mm}
\end{align*}$$

$$\begin{align*}
\text{Distance between the central maximum and the fifth bright fringe} &= 2 \times (\text{distance between consecutive fringes}) \times \text{number of fringes} \
&= 2 \times 0.65 \, \text{mm} \times 5 \
&= 6.5 \, \text{mm}
\end{align*}$$

Finally, to find the distance between the third dark fringe and the fifth bright fringe, we subtract the distance between the central maximum and the third dark fringe from the distance between the central maximum and the fifth bright fringe:

$$\begin{align*}
\text{Distance between the third dark fringe and the fifth bright fringe} &= \text{Distance between the fifth bright fringe} - \text{Distance between the third dark fringe} \
&= 6.5 \, \text{mm} - 3.9 \, \text{mm} \
&= 2.6 \, \text{mm}
\end{align*}$$

Therefore, the distance between the third dark fringe and the fifth bright fringe on the same side of the central maxima is 2.6 mm.

Similar Physics Doubts

A single slits of width 0.1 mm is illuminated by parallel light of wavelength 6000 A and diffraction bands are observed on a screen 40 cm from the slit. The distance (in mm) of third dark band from the central bright band is : Correct answer is '7.2'. Can you explain this answer?

If the distance between the first maxima and fifth minima of a double slit pattern is 7 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength (in nm) of the light used is : Correct answer is '600'. Can you explain this answer?

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