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Friday, 9 October

JEE Main 2017 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 2 April 2017 (offline), Physics Q23

Question 23 of 90 in this shift, Physics question 23 of 30.

In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :
  1. (1)1.56 mm
  2. (2)7.8 mmOfficial answer
  3. (3)9.75 mm
  4. (4)15.6 mm

Official answer

Option 2

CBSE answer key (25/04/2017, used for result).

Same topic in other shifts

All Young's Double Slit questions
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  2. 4 Apr 2026, Shift 2 · Q50In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe…MediumNumerical value
  3. 5 Apr 2026, Shift 1 · Q39In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is 2.4 μ2.4\ \mum. If the experiment…EasySingle correct
  4. 5 Apr 2026, Shift 2 · Q42The maximum intensity in a Young's double slit experiment is I0I_0. Distance between the slits (dd) is 5λ5\lambda, where λ\lambda is the…MediumSingle correct
  5. 6 Apr 2026, Shift 1 · Q36In interference experiment the path difference between two interfering waves at a point AA on the screen is λ/3\lambda/3, where λ\lambda…EasySingle correct
  6. 6 Apr 2026, Shift 2 · Q45In a Young double slit experiment, the wavelength of incident light is 6000 Å, the separation between slits S1S_1 and S2S_2 is 5 cm and…MediumDiagram basedHas a figure

Question text from the official JEE Main paper published by NTA (CBSE ran the 2017 exam); answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.