In YDSE, find the thickness of a glass slab (μ=1.5) which should be placed before the upper slit S1 so that the central maximum now lies at a point where 5 th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is 5000 mathringA.

Question

In YDSE, find the thickness of a glass slab (μ=1.5) which should be placed before the upper slit S1 so that the central maximum now lies at a point where 5 th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is 5000 mathringA. (A) 5 × 10-6 m (B) 3 × 10-6 m (C) 10 × 10-6 m (D) 5 × 10-5 m

Tardigrade Admin · Accepted Answer

According to the question, shift =5 fringe widths ⇒ ((μ-1) t D/d)=(5 λ D/d) ∴ t=(5 λ/μ-1)=(25000/1.5-1)=50,000 mathringA =5 × 10-6 m

Q. In YDSE, find the thickness of a glass slab $(μ = 1.5)$ which should be placed before the upper slit $S_{1}$ so that the central maximum now lies at a point where $5$ th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is $5000 \overset{˚}{A}$ .

$5 \times 1 0^{- 6} m$

$3 \times 1 0^{- 6} m$

$10 \times 1 0^{- 6} m$

$5 \times 1 0^{- 5} m$

According to the question,
shift $= 5$ fringe widths
$\Rightarrow \frac{( μ - 1 ) t D}{d} = \frac{5 λ D}{d}$
$∴ t = \frac{5 λ}{μ - 1} = \frac{25000}{1.5 - 1} = 50, 000 \overset{˚}{A}$
$= 5 \times 1 0^{- 6} m$

Q. In YDSE, find the thickness of a glass slab (μ=1.5) which should be placed before the upper slit S1​ so that the central maximum now lies at a point where 5 th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is 5000A˚.

5×10−6m

3×10−6m

10×10−6m

5×10−5m