In Young's double slit experiment, fringes are obtained using light of wavelength 4800 mathringA One slit is covered with a thin glass film of refractive index 1.4 and another slit is covered by a film of same thickness but refractive index 1.7 . By doing so, the central fringe is shifted to fifth bright fringe in the original pattern. The thickness of glass film is

Question

In Young's double slit experiment, fringes are obtained using light of wavelength 4800 mathringA One slit is covered with a thin glass film of refractive index 1.4 and another slit is covered by a film of same thickness but refractive index 1.7 . By doing so, the central fringe is shifted to fifth bright fringe in the original pattern. The thickness of glass film is (A) 2 × 10-3 mm (B) 4 × 10-3 mm (C) 6 × 10-3 mm (D) 8 × 10-3 mm

Tardigrade Admin · Accepted Answer

As per question, (μ1-1) t (D/d)-(μ2-1) (t D/d)=(5 λ D/d) [(1.7-1)-(1.4-1)] (t D/d)=(5 λ D/d) ∴ t=(5 λ/0.3)=(5 × 4800 × 10-10/0.3) =8 × 10-6 m =8 × 10-3 mm

$2 \times 1 0^{- 3} mm$

$4 \times 1 0^{- 3} mm$

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

$8 \times 1 0^{- 3} mm$

As per question,
$(μ_{1} - 1) t \frac{D}{d} - (μ_{2} - 1) \frac{t D}{d} = \frac{5 λ D}{d}$
$[(1.7 - 1) - (1.4 - 1)] \frac{t D}{d} = \frac{5 λ D}{d}$
$∴ t = \frac{5 λ}{0.3} = \frac{5 \times 4800 \times 1 0 ^{- 10}}{0.3}$
$= 8 \times 1 0^{- 6} m = 8 \times 1 0^{- 3} mm$

2×10−3mm

4×10−3mm

6×10−3mm

8×10−3mm

As per question, (μ1​−1)tdD​−(μ2​−1)dtD​=d5λD​ [(1.7−1)−(1.4−1)]dtD​=d5λD​ ∴t=0.35λ​=0.35×4800×10−10​ =8×10−6m=8×10−3mm

$2 \times 1 0^{- 3} mm$

$4 \times 1 0^{- 3} mm$

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

$8 \times 1 0^{- 3} mm$

As per question,
$(μ_{1} - 1) t \frac{D}{d} - (μ_{2} - 1) \frac{t D}{d} = \frac{5 λ D}{d}$
$[(1.7 - 1) - (1.4 - 1)] \frac{t D}{d} = \frac{5 λ D}{d}$
$∴ t = \frac{5 λ}{0.3} = \frac{5 \times 4800 \times 1 0 ^{- 10}}{0.3}$
$= 8 \times 1 0^{- 6} m = 8 \times 1 0^{- 3} mm$