KCET · Physics · Wave Optics
Young's double slit experiment gives interference fringes of width \(0.3 \mathrm{~mm}\). A thin glass plate made of material of refractive index \(1.5\) is kept in the path of light from one of the slits, then the fringe width becomes
- A zero
- B \(0.3 \mathrm{~mm}\)
- C \(0.45 \mathrm{~mm}\)
- D \(0.15 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(0.3 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
When a this glass plate of thickness \(t\) and refractive index \(\mu\) is introduced in one of the paths of the interfering waves then the path increases by \((\mu-1) \mathrm{t}\) and the whole pattern shifts by
\[
\mathrm{y}_{0}=\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}
\]
Shifting is towards the side in which the plate is introduced without any change in fringe width. Therefore, when a glass plate of refractive index \(1.5\) is kept in the path of light from one of the slits, only the fringes get shifted but the fringe width remains unchanged.
\[
\mathrm{y}_{0}=\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}
\]
Shifting is towards the side in which the plate is introduced without any change in fringe width. Therefore, when a glass plate of refractive index \(1.5\) is kept in the path of light from one of the slits, only the fringes get shifted but the fringe width remains unchanged.
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