| ⇦ | 
| ⇨ |
In Young’s double slit experiment, when wavelength used is 6000Å and the screen is 40 cm from the slits, the fringes are 0.012 cm wide. What is the distance between the slits?
Options
(a) 0.24 cm
(b) 0.2 cm
(c) 2.4 cm
(d) 0.024 cm
Correct Answer:
0.2 cm
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A Si and a Ge diode has identical physical dimensions. The bandgap in Si is larger
- A ball rolls off the top of stair way with a horizontal velocity of magnitude 1.8 m/s
- 1 Wb/m² equal to
- Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state
- If sink is at a temperature of -39⁰C and source at 0⁰C, then efficiency will be
Topics: Wave Optics
(101)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A Si and a Ge diode has identical physical dimensions. The bandgap in Si is larger
- A ball rolls off the top of stair way with a horizontal velocity of magnitude 1.8 m/s
- 1 Wb/m² equal to
- Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state
- If sink is at a temperature of -39⁰C and source at 0⁰C, then efficiency will be
Topics: Wave Optics (101)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply