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Step-by-Step Solution
Step 1: Understand the Given Data
• Slit separation, d = 0.3\,\text{mm} = 0.3 \times 10^{-3}\,\text{m} = 3 \times 10^{-4}\,\text{m}
• Screen distance, D = 1.5\,\text{m}
• Distance between the fourth bright fringes on both sides of the central bright fringe is 2.4\,\text{cm} = 2.4 \times 10^{-2}\,\text{m}
• We denote the wavelength of light by \lambda and the frequency by f .
• Speed of light, c = 3 \times 10^8\,\text{m/s} (assuming a standard value).
Step 2: Relate the Fringe Separation to Wavelength
In a Young’s double slit experiment, the separation between consecutive bright fringes (fringe width) is given by
\beta = \frac{\lambda D}{d}.
Since the distance between the fourth bright fringe on the positive side and the fourth bright fringe on the negative side is given as 2.4 \times 10^{-2}\,\text{m} , the total separation covers 8 fringe widths (from -4^\text{th} to +4^\text{th} ). Hence,
8 \beta = 2.4 \times 10^{-2}\,\text{m}.
Step 3: Express \beta in Terms of \lambda
Substitute \beta = \frac{\lambda D}{d} into the above equation:
8 \times \frac{\lambda D}{d} = 2.4 \times 10^{-2}.
Step 4: Solve for Wavelength \lambda
Rearrange to find \lambda :
\frac{8 \lambda D}{d} = 2.4 \times 10^{-2}
\quad \Longrightarrow \quad
\lambda = \frac{2.4 \times 10^{-2} \times d}{8 \, D}.
Plug in d = 3 \times 10^{-4}\,\text{m} and D = 1.5\,\text{m} :
\lambda = \frac{2.4 \times 10^{-2} \times (3 \times 10^{-4})}{8 \times 1.5}.
Carry out the multiplication and division carefully to get
\lambda = 6 \times 10^{-7}\,\text{m}.
Step 5: Determine the Frequency of Light
We know the relationship between wavelength \lambda , frequency f , and speed of light c :
c = \lambda \, f \quad \Longrightarrow \quad f = \frac{c}{\lambda}.
Substitute \lambda = 6 \times 10^{-7}\,\text{m} and c = 3 \times 10^8\,\text{m/s} :
f = \frac{3 \times 10^8}{6 \times 10^{-7}} = 5 \times 10^{14}\,\text{Hz}.
Final Answer
The frequency of the light used in the experiment is
5 \times 10^{14}\,\text{Hz}.
