JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get
a detailed list of all topics that are important in cracking the entrance
examination. JEE Advanced syllabus for Physics has been designed in such a
way that it offers very practical and application-based learning to
further make it easier for students to understand every concept or topic
by correlating it with day-to-day experiences. In comparison to the other
two subjects, the syllabus of JEE Advanced for physics is developed in
such a way so as to test the deep understanding and application of
concepts.
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Q1. Interference fringes were produced using light in a double-slit experiment. When a mica sheet of uniform thickness and refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some distance. This distance is equal to the width of 30 interference bands if light of wavelength 4800 is used. The thickness (in μm) of mica is
Q2.In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is
Q3. High-quality camera lenses are often coated to prevent reflection. A lens has an optical index of refraction 1.72 and a coating with an optical index of refraction 1.31. For near normal incidence, the minimum thickness of the coating to prevent reflection for wavelength of 5.3 × 10-7m is
Q4. A Young’s double-slit experiment is conducted in water (𝜇1) as shown in figure, and a glass plate of thickness 𝑡 and refractive index 𝜇2 is placed in the path of 𝑆2. Find the magnitude of the optical path difference at ‘𝑂’
Q5.Figure shows a wavefront 𝑃 passing through two systems 𝐴 and 𝐵, and emerging as 𝑄 and then as 𝑅. The systems 𝐴 and 𝐵 could, respectively, be
Q6. In a Young’s double-slit experiment, 30 fringes are obtained in the field of view of the observing telescope, when the wavelength of light used in 4000 Å. If we use monochromatic light of wavelength 6000 Å, the number of fringes obtained in the same field of view is
Q7.In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m. The minimum distance between two successive regions of complete darkness is
Q8.In YDSE, coherent monochromatic light having wavelength 600 nm falls on the slits. First-order bright fringe is at 4.84 mm from central maxima. Determine the wavelength for which the first-order dark fringe will be observed at the same location on screen. (Take 𝐷 = 3 m)
Q9.The YDSE apparatus is as shown in figure. The condition for point 𝑃 to be a dark fringe is
(𝑙 = wavelength of light waves)
Q10. In a Young’s double-slit experiment, 𝜆 = 500 nm, 𝑑 = 1 nm, and 𝐷 = 1 m. The minimum distance from the central maximum for which the intensity is half of the maximum intensity is