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Question: A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thi...

A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away.

(a) What would be the fringe-width ?

(b) At what distance from the centre will the first maximum be located?

Answer

(a) Fringe-width: 4.92×10−44.92 \times 10^{-4} m (b) Distance of the first maximum from the center: 1.30×10−21.30 \times 10^{-2} m

Explanation

Solution

The given problem is about Young's Double Slit Experiment with thin films placed in front of the slits.

Given parameters:

Thickness of the strips, t=0.50t = 0.50 mm =0.50×10−3= 0.50 \times 10^{-3} m Separation between the slits, d=0.12d = 0.12 cm =0.12×10−2= 0.12 \times 10^{-2} m Refractive index of mica, μ1=1.58\mu_1 = 1.58 Refractive index of polysterene, μ2=1.55\mu_2 = 1.55 Wavelength of light, λ=590\lambda = 590 nm =590×10−9= 590 \times 10^{-9} m Distance of the screen from the slits, D=1D = 1 m

(a) Fringe-width (β\beta):

The fringe-width in a double-slit experiment is given by the formula β=λDd\beta = \frac{\lambda D}{d}. The presence of thin films in front of the slits shifts the entire interference pattern but does not change the spacing between the fringes, i.e., the fringe-width, provided the medium between the slits and the screen is uniform.

Substituting the given values: β=(590×10−9 m)×(1 m)(0.12×10−2 m)\beta = \frac{(590 \times 10^{-9} \text{ m}) \times (1 \text{ m})}{(0.12 \times 10^{-2} \text{ m})} β=590×10−90.12×10−2 m=5900.12×10−7 m\beta = \frac{590 \times 10^{-9}}{0.12 \times 10^{-2}} \text{ m} = \frac{590}{0.12} \times 10^{-7} \text{ m} β=5900012×10−7 m=147503×10−7 m\beta = \frac{59000}{12} \times 10^{-7} \text{ m} = \frac{14750}{3} \times 10^{-7} \text{ m} β≈4916.67×10−7 m=4.91667×10−4 m\beta \approx 4916.67 \times 10^{-7} \text{ m} = 4.91667 \times 10^{-4} \text{ m} Rounding to three significant figures, β≈4.92×10−4\beta \approx 4.92 \times 10^{-4} m. In mm, β≈0.492\beta \approx 0.492 mm.

(b) Distance of the first maximum from the center (y1y_1):

When thin films of thickness tt and refractive indices μ1\mu_1 and μ2\mu_2 are placed in front of the two slits, the optical path difference at a point on the screen at a distance yy from the center is given by ΔL=ydD+(μ2−μ1)t\Delta L = \frac{yd}{D} + (\mu_2 - \mu_1)t. Here, we assume the mica strip is in front of slit S1 and the polysterene strip is in front of slit S2, and yy is measured from the center of the screen, positive upwards. The term (μ2−μ1)t(\mu_2 - \mu_1)t represents the initial path difference introduced by the films at the center of the screen (y=0y=0).

The central maximum (n=0n=0) occurs when ΔL=0\Delta L = 0. Let y0y_0 be the position of the central maximum. y0dD+(μ2−μ1)t=0\frac{y_0 d}{D} + (\mu_2 - \mu_1)t = 0 y0=−(μ2−μ1)tDd=(μ1−μ2)tDdy_0 = -\frac{(\mu_2 - \mu_1)t D}{d} = \frac{(\mu_1 - \mu_2)t D}{d}

Substituting the given values: μ1−μ2=1.58−1.55=0.03\mu_1 - \mu_2 = 1.58 - 1.55 = 0.03 y0=(0.03)×(0.50×10−3 m)×(1 m)(0.12×10−2 m)y_0 = \frac{(0.03) \times (0.50 \times 10^{-3} \text{ m}) \times (1 \text{ m})}{(0.12 \times 10^{-2} \text{ m})} y0=0.015×10−30.12×10−2 m=0.0150.12×10−1 my_0 = \frac{0.015 \times 10^{-3}}{0.12 \times 10^{-2}} \text{ m} = \frac{0.015}{0.12} \times 10^{-1} \text{ m} y0=15120×10−1 m=18×10−1 m=0.125×10−1 m=0.0125 my_0 = \frac{15}{120} \times 10^{-1} \text{ m} = \frac{1}{8} \times 10^{-1} \text{ m} = 0.125 \times 10^{-1} \text{ m} = 0.0125 \text{ m} y0=1.25×10−2y_0 = 1.25 \times 10^{-2} m =1.25= 1.25 cm =12.5= 12.5 mm. This is the position of the central maximum relative to the original center.

The first maximum (n=1n=1) occurs when ΔL=1λ\Delta L = 1\lambda. Let y1y_1 be the position of the first maximum. y1dD+(μ2−μ1)t=λ\frac{y_1 d}{D} + (\mu_2 - \mu_1)t = \lambda y1dD=λ−(μ2−μ1)t=λ+(μ1−μ2)t\frac{y_1 d}{D} = \lambda - (\mu_2 - \mu_1)t = \lambda + (\mu_1 - \mu_2)t y1=λDd+(μ1−μ2)tDdy_1 = \frac{\lambda D}{d} + \frac{(\mu_1 - \mu_2)t D}{d} The first term is the standard position of the first maximum without the films relative to the original center (β\beta). The second term is the shift in the central maximum (y0y_0). So, y1=β+y0y_1 = \beta + y_0.

Using the calculated values: β≈4.91667×10−4\beta \approx 4.91667 \times 10^{-4} m y0=0.0125y_0 = 0.0125 m y1=0.0125 m+4.91667×10−4 my_1 = 0.0125 \text{ m} + 4.91667 \times 10^{-4} \text{ m} y1=0.0125 m+0.000491667 my_1 = 0.0125 \text{ m} + 0.000491667 \text{ m} y1=0.012991667 my_1 = 0.012991667 \text{ m} Rounding to three significant figures, y1≈0.0130y_1 \approx 0.0130 m. In cm, y1≈1.30y_1 \approx 1.30 cm. In mm, y1≈13.0y_1 \approx 13.0 mm.

Final Answers: (a) The fringe-width is approximately 4.92×10−44.92 \times 10^{-4} m (or 0.4920.492 mm). (b) The first maximum is located at a distance of approximately 0.01300.0130 m (or 1.301.30 cm or 13.013.0 mm) from the center.

(a) The fringe width in YDSE depends on the wavelength of light, slit separation, and screen distance. The thin films shift the pattern but do not alter the fringe width. (b) The thin films introduce an initial optical path difference (μ1−μ2)t(\mu_1 - \mu_2)t. The central maximum shifts to a position where this initial path difference is cancelled by the geometric path difference. The first maximum is located one fringe width away from the central maximum position in the direction of increasing optical path difference from the slit with lower refractive index film (polysterene).