In a Young's double-slit experiment, the slit separation is d = 0.20 mm, the screen distance is D = 1.0 m, and light of wavelength lambda = 600 nm is used. What is the fringe width (fringe separation beta)?
Answer and explanation
Correct answer: C. 3.0 millimeters
- Step 1: Fringe width formula: beta = lambda D / d. Step 2. lambda = 600 x 10^-9 m, D = 1.0 m, d = 0.20 x 10^-3 m = 2.0 x 10^-4 m. Step 3. beta = (600 x 10^-9 1.0) / (2.0 x 10^-4) = 3.0 x 10^-3 m = 3.0 mm. Answer: 3.0 millimeters. Why not the others: 0.30 mm has a factor of 10 error; 1.2 mm results from d = 0.5 mm.
Practise more Waves, Oscillations and Optics.
Free questions with an explanation for every one, and full timed MDCAT Pakistan mock tests.
More Waves, Oscillations and Optics questions
- A simple pendulum of length L has a period T on Earth. If its length is quadrupled (4 L) and it is taken to the Moon where acceleration due…
- An ambulance siren emits a frequency of 600 Hz while approaching a stationary observer at a speed of 30 m/s. What frequency does the observe…
- What is the critical angle for total internal reflection at a boundary between glass (refractive index n1 = 1.50) and water (n2 = 1.33)?
- An object is placed 15 cm in front of a convex (converging) lens of focal length 10 cm. What are the image distance and image characteristic…
- For a pipe of length L open at both ends, what is the wavelength of the fundamental (first harmonic) standing wave?
- In simple harmonic motion, at what displacement from the equilibrium position is the kinetic energy equal to the potential energy?