In a Young’s double-slit experimen | Class 12 Physics Chapter Wave Optics, Wave Optics NCERT Solutions

Welcome to the NCERT Solutions for Class 12 Physics - Chapter Wave Optics. This page offers a step-by-step solution to the specific question from Exercise 1, Question 4: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 4:

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer:

Here it is given that,

Distance between the slits, d = 0.28 mm = 0.28 × 10 -3 m

Distance between the slits and the screen, D = 1.4 m

Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 × 10 -2 m

For constructive interference, the distance between the two fringes is given by relation: u = nλ D/d

where, n = Order of fringes

wavelength of the light can be given as: λ = ud/nD = 1.2x10-2x0.28x10-3/4x1.4 = 6x10-7 = 600 nm

Hence, the wavelength of the light is 6 x 10 -7 m.


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  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Wave Optics.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 4: In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is plac....