Previous year questions: Wave Optics
Wave Optics28 min readLesson 6 of 6
Q1 (CBSE Board 2022, 3 marks)
In a single slit diffraction experiment, draw the intensity distribution and obtain the condition for the first secondary maximum. Answer: The central maximum is the widest band and the intensity falls quickly on both sides. The minima occur where a sine theta equals m lambda. The first minimum is at a sine theta equal to lambda, so the first secondary maximum lies near a sine theta equal to three lambda by two. Thus theta equals the inverse sine of three lambda divided by twice the slit width. The width of the central maximum is the distance between the two first minima, so it equals twice D lambda divided by the slit width. The distribution is a central band with rapidly fading side bands.
Q2 (CBSE Board 2022, 2 marks)
In Young's double slit experiment, state how the fringe width changes when the screen is moved closer to the slits and when the separation between the slits is increased. Answer: Fringe width equals lambda D divided by d, so it is proportional to the screen distance D and inversely proportional to the slit separation d. Moving the screen closer decreases the fringe width. Increasing the separation between the slits also decreases the fringe width. The intensity of the bright fringes is not changed by these geometric changes. A numeric check: for a wavelength of 600 nm, a screen distance of 1.5 m and a slit separation of 0.3 mm, the width works out to 3 mm, and moving the screen to 1.0 m reduces it to 2 mm. The fringe width is proportional to the screen distance and inversely proportional to the slit separation.
Q3 (CBSE Board 2022, 2 marks)
Two waves meeting at a point on a screen have a path difference of 2 lambda by 3. State whether the point is bright or dark, and repeat for a path difference of 2 lambda. Answer: A path difference of 2 lambda by 3 is a fraction of a wavelength, not a whole number and not an odd multiple of half a wavelength, so the point is neither a perfect bright nor a perfect dark fringe. A path difference of 2 lambda is a whole number of wavelengths, so the point is a bright fringe. A path difference of two thirds of a wavelength corresponds to a phase difference of four pi by three, which lies between a bright and a dark condition. Only whole-number or half-odd multiples of the wavelength give pure bright or dark points.
Q4 (CBSE Board 2023, 5 marks)
Write two ways in which an interference pattern and a diffraction pattern differ. List two quantities that set the spacing of fringes. In an arrangement where the two openings are 100 wavelengths apart and the viewing screen sits 50 cm from them, obtain the angular gap between neighbouring maxima and the linear gap on the screen. Answer: Interference fringes have a common width and the intensity falls off slowly, while diffraction has a central band twice as wide and much brighter than the side bands. The spacing is set by the wavelength and by the ratio of the screen distance to the opening separation. The angular gap is lambda divided by the separation, equal to 0.01 radian. The linear gap is the screen distance times this angle, giving 0.005 m, that is 0.5 cm. The five mark part also expects the intensity distribution to be drawn with a wide central maximum and equally spaced minima. The two answers are consistent because the screen distance times the angle gives the linear gap.
Q5 (CBSE Board 2024, 2 marks)
Two waves of intensities I1 and I2 superpose at a point with a phase difference phi. Obtain the expression for the resultant intensity. Answer: The resultant amplitude is found by adding the two amplitude vectors with the phase angle between them. Squaring gives the resultant intensity as I1 plus I2 plus twice the square root of I1 I2 times cosine phi. For equal intensities this becomes 4I cosine squared phi by 2, so the intensity varies between zero and four times the intensity of one wave. For unequal intensities the minima do not reach zero, and the contrast is described by the ratio of the two amplitudes. Setting the two intensities equal gives the maximum value of four times the individual intensity.
Q6 (CBSE Board 2024, 2 marks)
State the effect on the interference pattern in Young's double slit experiment when the source slit is moved closer to the plane of the two slits, and when the separation between the two slits is increased. Answer: Moving the source slit closer to the plane of the slits does not change the fringe width, because the width depends on the distance of the screen from the slits and on their separation. Increasing the separation between the slits decreases the fringe width, since the width is inversely proportional to the slit separation. In both cases the fringes remain equally spaced. Covering one slit removes the interference and leaves only the diffraction pattern of the open slit on the screen. Fringe width does not depend on how far the source slit is from the double slit.
Q7 (CBSE Board 2025, 3 marks)
A double slit set-up is first placed in water and then the water is replaced by a transparent liquid of refractive index greater than that of water. State the effect on the speed, the frequency and the fringe width of the light. Answer: The speed of light in the liquid is smaller than in water because the refractive index is larger. The frequency is set by the source and does not change when the medium changes. The wavelength becomes smaller, so the fringe width, which is proportional to wavelength, also decreases. The wavelength in the liquid is lambda divided by n, so the fringe width changes by the same factor as the wavelength. The frequency is fixed by the source, so only the wavelength changes.
Q8 (CBSE Board 2025, 2 marks)
In a single slit diffraction experiment the slit width is 0.55 mm and the wavelength of light is 550 nm. Find the angle of diffraction for the first secondary maximum. Answer: The first secondary maximum is near a sine theta equal to three lambda by two. Substituting the values gives sine theta equal to 1.5 times 550 nanometres divided by 0.55 millimetres, which is 1.5 times 10 to the minus 3. For such a small angle the angle in radians is also 1.5 times 10 to the minus 3, which is about 0.086 degrees. The second secondary maximum lies near a sine theta equal to five lambda by two, at an angle about three times the first one. For a small angle the sine and the angle in radians are numerically equal.
Q9 (CBSE Board 2022, 2 marks)
State the two conditions for two sources of light to be coherent, and explain why two independent sodium lamps cannot produce a sustained interference pattern. Answer: Coherent sources must emit light of the same frequency and must maintain a constant phase difference. Two independent sodium lamps contain many atoms emitting randomly and independently, so the phase difference between their waves changes rapidly with time. The pattern shifts continuously and averages out to uniform illumination. The random phases spread the interference pattern over time, so the screen shows uniform brightness instead of steady fringes. This is a statement-based question, so the marks come from the reasoning and not from arithmetic.
Q10 (CBSE Board 2022, 2 marks)
A Young's double slit experiment is performed with light of wavelength 520 nm and then with an unknown wavelength, keeping the geometry the same. If the fourth bright fringe of the unknown wavelength falls on the same point as the fifth bright fringe of the known light, find the unknown wavelength. Answer: For the same geometry the position of the nth bright fringe is proportional to n times the wavelength. Equating the two positions gives 4 times lambda unknown equal to 5 times 520 nm. Therefore lambda unknown equals 5 times 520 divided by 4, which is 650 nm. Using the same geometry is what makes the fringe positions proportional to the order and to the wavelength alone. The ratio of the orders is enough to find the unknown wavelength.