If the wavelength of light used in Young's experiment is 600 nm, what is the fringe width when the distance between the slits is 0.1 mm and the distance to the screen is 2 m?
Practice Questions
1 question
Q1
If the wavelength of light used in Young's experiment is 600 nm, what is the fringe width when the distance between the slits is 0.1 mm and the distance to the screen is 2 m?
0.03 mm
0.06 mm
0.12 mm
0.15 mm
Fringe width (β) = (λD)/d = (600 x 10^-9 * 2)/(0.1 x 10^-3) = 0.012 mm = 0.06 mm.
Questions & Step-by-step Solutions
1 item
Q
Q: If the wavelength of light used in Young's experiment is 600 nm, what is the fringe width when the distance between the slits is 0.1 mm and the distance to the screen is 2 m?
Solution: Fringe width (β) = (λD)/d = (600 x 10^-9 * 2)/(0.1 x 10^-3) = 0.012 mm = 0.06 mm.
Steps: 10
Step 1: Identify the given values from the problem. The wavelength of light (λ) is 600 nm, the distance between the slits (d) is 0.1 mm, and the distance to the screen (D) is 2 m.
Step 2: Convert the units to be consistent. Convert 600 nm to meters: 600 nm = 600 x 10^-9 m. Convert 0.1 mm to meters: 0.1 mm = 0.1 x 10^-3 m.
Step 3: Write the formula for fringe width (β): β = (λD) / d.
Step 4: Substitute the values into the formula. β = (600 x 10^-9 m * 2 m) / (0.1 x 10^-3 m).
Step 5: Calculate the numerator: 600 x 10^-9 m * 2 m = 1200 x 10^-9 m.
Step 6: Calculate the denominator: 0.1 x 10^-3 m = 0.0001 m.
Step 7: Divide the numerator by the denominator: β = (1200 x 10^-9 m) / (0.0001 m).
Step 8: Perform the division: β = 0.012 m.
Step 9: Convert the result back to mm: 0.012 m = 12 mm.
Step 10: The final answer for fringe width is 12 mm.
