Question

In YDSE a glass slab of thickness \(8\,\mu m\) is introduced in front of a slit. If central maxima shifts to the position of 4th minima, then find refractive index of glass slab \((\lambda = 500\,\text{nm})\):

• A. \( \mu = 1.11 \)
• B. \( \mu = 1.22 \)
• C. \( \mu = 1.32 \)
• D. \( \mu = 2.2 \)

Answer 1

The Correct Option is B

Concept: When a thin glass slab of thickness \(t\) and refractive index \(\mu\) is inserted in front of one slit in YDSE, the central fringe shifts by \[ y_0 = (\mu - 1)t\frac{D}{d} \] Position of minima in YDSE is \[ y_n = \left(n+\frac{1}{2}\right)\frac{D\lambda}{d} \] Step 1: Central maxima shifts to the position of 4th minima.} For minima, \[ y_n=\left(n+\frac{1}{2}\right)\frac{D\lambda}{d} \] 4th minima corresponds to \(n=3\) \[ y_1=\left(3+\frac{1}{2}\right)\frac{D\lambda}{d} \] \[ y_1=\frac{7}{2}\frac{D\lambda}{d} \] Since the central maxima shifts to this position, \[ y_0 = y_1 \]
Step 2: Equate shift expression.} \[ (\mu-1)t\frac{D}{d}=\frac{7}{2}\frac{D\lambda}{d} \] Cancel \( \frac{D}{d} \): \[ (\mu-1)t=\frac{7}{2}\lambda \]
Step 3: Substitute given values.} \[ t=8\times10^{-6}\,m \] \[ \lambda=500\times10^{-9}\,m \] \[ (\mu-1)=\frac{7}{2}\times\frac{500\times10^{-9}}{8\times10^{-6}} \] \[ (\mu-1)=3.5\times62.5\times10^{-3} \] \[ (\mu-1)=0.218 \] \[ \mu=1.218 \approx 1.22 \] Final Result \[ \mu \approx 1.22 \]