If the wavelength of light used in an interference experiment is 500 nm, what is
Practice Questions
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
Questions & Step-by-Step Solutions
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?
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.
Interference Patterns – The question tests understanding of the interference of light waves and the calculation of fringe separation in a double-slit experiment.
Wavelength and Distance Relationships – It assesses the ability to apply the formula for fringe separation, which relates wavelength, distance to the screen, and slit separation.
