In the Young's double slit experiment, the spacing between two slits is 0.1 mm. If the screen is kept at a distance of 1.0 m from the slits and the wavelength of light is 5000 Å, then the fringe width is (a) 1.0 cm (b) 1.5 cm (c) 0.5 cm (d) 2.0 cm
Added by Dev D.
Step 1
1 mm = 0.1 x 10^-3 m - Distance from slits to screen (L) = 1.0 m - Wavelength of light (λ) = 5000 Å = 5000 x 10^-10 m Show more…
Show all steps
Your feedback will help us improve your experience
Kamlesh Goyal and 79 other Physics 103 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In Young's double slit experiment, the distance between the two slits is 0.1 mm and the wavelength of light used is m 4 *10^-7 . If the width of the fringe on the screen is 4 mm, the distance between screen and slit is (a) 0.1 mm (b) 1 cm (c) 0.1 cm (d) 1 m
Kamlesh G.
If in a Young's double slit experiment, the slit width is 3 cm, the separation between slits and screen is 70cm and wavelength of light is 1000 A , then fringe width will be ( mu = 1.5) A. 2.3 × 10^-3 cm B. 2.3 × 10^-4 m C. 2.3 × 10^-5 cm D. 2.3 × 10^-6 m
Sri K.
In Young's double slit experiment with monochromatic light of wavelength $600 \mathrm{~nm}$, the distance between slits is $10^{-3} \mathrm{~m}$. For changing fringe width by $3 \times 10^{-5} \mathrm{~m}$ : (a) the screen is moved away from the slits by $5 \mathrm{~cm}$ (b) the screen is moved by $5 \mathrm{~cm}$ towards the slits (c) the screen is moved by $3 \mathrm{~cm}$ towards the slits (d) both (a) and (b) are correct
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD