If the wavelength of light used in an interference experiment is 500 nm, what is the fringe separation when the screen is placed 2 m away from the slits separated by 0.1 mm?
Practice Questions
1 question
Q1
If the wavelength of light used in an interference experiment is 500 nm, what is the fringe separation when the screen is placed 2 m away from the slits separated by 0.1 mm?
0.01 m
0.025 m
0.05 m
0.1 m
Fringe separation β = λD/d = (500 x 10^-9 m)(2 m)/(0.1 x 10^-3 m) = 0.01 m.
Questions & Step-by-step Solutions
1 item
Q
Q: If the wavelength of light used in an interference experiment is 500 nm, what is the fringe separation when the screen is placed 2 m away from the slits separated by 0.1 mm?
Solution: Fringe separation β = λD/d = (500 x 10^-9 m)(2 m)/(0.1 x 10^-3 m) = 0.01 m.
Steps: 8
Step 1: Identify the given values from the problem. The wavelength of light (λ) is 500 nm, the distance to the screen (D) is 2 m, and the distance between the slits (d) is 0.1 mm.
Step 2: Convert the units to meters for consistency. 500 nm = 500 x 10^-9 m and 0.1 mm = 0.1 x 10^-3 m.
Step 3: Use the formula for fringe separation, which is β = λD/d.
Step 4: Substitute the values into the formula: β = (500 x 10^-9 m)(2 m) / (0.1 x 10^-3 m).
Step 5: Calculate the numerator: (500 x 10^-9 m) * (2 m) = 1000 x 10^-9 m = 1 x 10^-6 m.
Step 6: Calculate the denominator: 0.1 x 10^-3 m = 0.0001 m.
Step 7: Divide the numerator by the denominator: β = (1 x 10^-6 m) / (0.0001 m) = 0.01 m.
Step 8: Conclude that the fringe separation is 0.01 m.
