The maximum number of possible interference maxima for slit-seperation equal to

⇦

⇨

The maximum number of possible interference maxima for slit-seperation equal to twice the wavelength, in young’s double slit experiment is

Options

(a) Infinite
(b) five
(c) three
(d) zero

Correct Answer:

five

Explanation:

For interference maxima, d sin θ = nλ

⇒ 2λ sin θ = nλ ⇒ sin θ = n / 2

sin θ can have values between 0 and ±1.

Hence n can be (-2, -1, 0, +1, +2) or five values.

The possible maxima are five.

Related Questions:

1. A wire of one metre length carries a constant current. The wire is bent to form
2. The numebr 0 (zero) is required for
3. A rectangular copper coil is placed in a uniform magnetic field of induction 40 mT
4. The ratio of amplitude of magnetic field to the amplitude of electric field for
5. In a circuit, 5% of total current passes through a galvanometer. If resistance

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