Question

Monochromatic light falls on two narrow slits that are 0.0190 mm apart. A first order fringe is 16 mm from the central line. The screen (back wall) is 0.700 m from the slits. What is the wavelength of the light?

Group of answer choices

2.01×10−7 m

4.34×10−7 m

6.68×10−7 m

5.11×10−7 m

Answer 1

To find the wavelength of the light, we can use the formula for the position of the interference fringes in a double-slit experiment:

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

where:
- \( y \) is the distance from the central maximum to the m-th order maximum (in meters).
- \( m \) is the order of the fringe (for first order, \( m = 1 \)).
- \( \lambda \) is the wavelength of the light (in meters).
- \( L \) is the distance from the slits to the screen (in meters).
- \( d \) is the distance between the slits (in meters).

Given:
- \( y = 16 \, \text{mm} = 0.016 \, \text{m} \)
- \( m = 1 \) (first order fringe)
- \( L = 0.700 \, \text{m} \)
- \( d = 0.0190 \, \text{mm} = 0.0000190 \, \text{m} \)

We want to find \( \lambda \). Rearranging the formula for \( \lambda \):

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

Substituting the values into the equation:

\[
\lambda = \frac{0.016 \, \text{m} \cdot 0.0000190 \, \text{m}}{1 \cdot 0.700 \, \text{m}}
\]

Calculating the numerator:

\[
0.016 \, \text{m} \cdot 0.0000190 \, \text{m} = 0.000000304 \, \text{m}^2
\]

Then dividing by \( L \):

\[
\lambda = \frac{0.000000304 \, \text{m}^2}{0.700 \, \text{m}} = 0.0000004342857 \, \text{m}
\]

Thus,

\[
\lambda \approx 4.34 \times 10^{-7} \, \text{m}
\]

The closest choice available is:

**4.34×10−7 m**