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NBT1 2016

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NBT1 2016

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krisretro1
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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THE OPEN UNIVERSITY OF SRI LANKA

BACHELOR OF SCIENCE DEGREE PROGRAMME – LEVEL 04


NO BOOK TEST-1 (2015/2016)
OPTICS –PYU 2164
Duration: ONE HOUR (01 h)
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Date: 2nd April 2016 Time: 4.00 pm – 5.00 pm
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Answer All (02) questions

1.(a) Obtain an expression for the resultant amplitude at a point in the region of superposition
of two waves, y1= a1 sin t and y2= a2 sin ( t + ). Where is the phase difference
between the waves at that point (The other symbols have their usual meaning). Hence,
deduce an expression for the resultant intensity at that point.

(b). Two beams of coherent light of wavelength 5000 Å reaches a point with individual
intensities of 1.44 and 4.00 units. After they interfere, the intensity is 0.90 units.
Calculate the lowest phase difference of the beams reach at that point.

(c). Describe the formation of fringes by interference of light using two parallel slits in
Young’s double slit experiment. Obtain an expression for the fringe width. What is the
effect of increasing the width of the slit?

(d). In Young’s double slit experiment, the separation of the slit is 1.9 mm and the fringe
spacing is 0.31 mm at a distance of 1 m from the slits. Calculate the wavelength of light.

2. (a). It is often considered only two rays to discuss the interference phenomenon of a thin
films either due to the multiple reflections or the transmitted light. Obtain an expression
for a maximum (constructive interference) due to the interference of two rays reflected
from the thin film.

(b). A parallel beam of monochromatic light of wavelength 5500 Å falls normally on a


thin film of oil of refractive index 1.45. Calculate the minimum thickness of the film so
that the light is reflected strongly.

(c). Draw a neat, labeled diagram of experimental setup of the Newton’s rings. Explain
with the necessary theory, how it can be used to determining the wavelength of sodium
light.

(d). In a Newton’s ring experiment the diameter of the 10th dark ring changes from 0.40 cm
to 1.27 cm when a liquid is introduced between the lens and the glass plate. Calculate
the refraction index of the liquid.

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