Monochromatic light falls on two narrow slits that are 0.0190 mm apart. A first order fringe is 15 mm from the central line. The screen (back wall) is 0.600 m from the slits. What is the wavelength of the light?

Question

Group of answer choices

2.01×10−7 m

6.68×10−7 m

4.75×10−7 m

5.11×10−7 m

Answer 1

To find the wavelength of the monochromatic light, we can use the double-slit interference formula:

\[ y = \frac{m \lambda L}{d} \]

where:

• \( y \) is the distance from the central maximum to the m-th order maximum (15 mm for the first order, so \( y = 0.015 , \text{m} \))
• \( m \) is the order of the fringe (1 for first order)
• \( \lambda \) is the wavelength of the light (what we want to find)
• \( L \) is the distance from the slits to the screen (0.600 m)
• \( d \) is the distance between the two slits (0.0190 mm = \( 0.0190 \times 10^{-3} , \text{m} \))

Now, let's rearrange the formula to solve for \( \lambda \):

\[ \lambda = \frac{y d}{m L} \]

Substituting in the values:

\[ \lambda = \frac{0.015 \times 0.0190 \times 10^{-3}}{1 \times 0.600} \]

Calculating the numerator:

\[ 0.015 \times 0.0190 \times 10^{-3} = 0.000000285 , \text{m} \]

Now for the entire calculation:

\[ \lambda = \frac{0.000000285}{0.600} = 0.000000475 , \text{m} = 4.75 \times 10^{-7} , \text{m} \]

Thus, the wavelength of the light is:

\[ \lambda = 4.75 \times 10^{-7} , \text{m} \]

The correct answer from the given choices is:

4.75×10−7 m