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JKCET Young's Double Slit — practice questions

Explore 3 free Young's Double Slit questions in JKCET Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Young's Double Slit questions and exercises with answers

Showing questions 1–3 of 3.

In Young's double-slit experiment, the fringe width $w$ is given by:Read the answer and worked solution →In Young's experiment, if the entire setup is immersed in a medium of refractive index $n$, the fringe width:Read the answer and worked solution →Two slits 0.5 mm apart are illuminated with light of 600 nm. The fringes are observed on a screen 1 m away. The fringe width is:Read the answer and worked solution →

Worked practice questions

In Young's double-slit experiment, the fringe width $w$ is given by:

  1. $w = \lambda d / D$
  2. $w = \lambda D / d$
  3. $w = D / (\lambda d)$
  4. $w = (\lambda + D)/d$
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Correct answer: B. $w = \lambda D / d$

Fringe width (separation between adjacent maxima or minima) $w = \lambda D/d$, where $D$ = screen-slit distance, $d$ = slit separation, $\lambda$ = wavelength. Bigger $\lambda$ or $D$, or smaller $d$ ⇒ wider fringes.

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In Young's experiment, if the entire setup is immersed in a medium of refractive index $n$, the fringe width:

  1. Stays the same
  2. Increases by factor $n$
  3. Decreases by factor $n$ (since $\lambda$ shortens)
  4. Becomes zero
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Correct answer: C. Decreases by factor $n$ (since $\lambda$ shortens)

Wavelength in medium $\lambda' = \lambda/n$, so fringe width $w' = \lambda' D/d = w/n$. **Hence immersion in water (n=1.33) shrinks fringes by ~25%.**

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Two slits 0.5 mm apart are illuminated with light of 600 nm. The fringes are observed on a screen 1 m away. The fringe width is:

  1. 0.6 mm
  2. 1.2 mm
  3. 1.5 mm
  4. 3 mm
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Correct answer: B. 1.2 mm

We have $w = \lambda D/d$. That is $w = (600 \times 10^{-9})(1)/(0.5 \times 10^{-3})$. Then $w = 1.2 \times 10^{-3}$. So m = 1.2 mm.

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