Question

In YDSE, slit separation $d = 2$ mm, distance between slits and screen $D = 10$ m. Wavelength of light is $\lambda = 6000$\AA. If intensity of light through each slit is $I_0$, find intensity at point directly in front of one of the slits.

• A. $4I_0$
• B. Zero
• C. $I_0$
• D. $2I_0$

Hint:

In YDSE, intensity depends on phase difference even if observation point is not at center.

Answer 1

The Correct Option is C

Step 1: Recall the intensity formula in YDSE

In Young’s Double Slit Experiment, the resultant intensity at any point on the screen is:

I = I₁ + I₂ + 2√(I₁I₂) cos φ

where φ is the phase difference corresponding to the path difference.

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Step 2: Consider the point directly in front of one slit

At a point directly in front of one slit (say slit S₁):

• Light from slit S₁ reaches normally.
• Light from slit S₂ does not contribute effectively at this point due to geometry.

Hence, there is no effective interference at this point.

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Step 3: Evaluate the intensity at this point

Since only one slit contributes:

I = I₀

There is no constructive or destructive interference here.

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Final Answer:

The intensity at the point directly in front of one slit is
I₀