Sign up now
to enroll in courses, follow best educators, interact with the community and track your progress.
Solved Examples on Young’s Double Slit Experiment (in Hindi)
169 plays

In this lecture you will get some very good numerical based on YDSE, it will help you in JEE. Let's see.

Kartikey Pandey
IIT BHU(B.Tech)| Super 30 Physics Faculty | FOUNDER of New era online coaching. (1,76,000+ Subscribers)

Unacademy user
Thanks a lot sir for making History understandable in easy language...You made a beautiful story and sometimes when a person tells story he/she misses out some facts and important points but I have cross checked and found that you didn't missed a single point...Awesome sir..thanks a lot...
sir please bring 2 more videos on solved examples of ydse they are very helpful
Very helpful.... some more important solved examples.. Please upload brewsters law and law of malus problems also which repeatedly comes in jee main..
Sir lecture for emi

  1. Monochromatic light of wavelength 5000 A is used in Y.D.S.E., with slit-width, d 1mm, distance between screen and slits, D 1 m. If intensity at the two slits are, 1-410, 1.-10, find (i) fringe width Important (ii) distance of 5th minima from the central maxima on the screen (ii) Intensity at y mm (iv) Distance of the 1000th maxima from the central maxima on the screen. (v) Distance of the 5000th maxima from the central maxima on the screen

  2. D 5000 x10-10x1 = 0.5 mm 1x10-3 AD 2d y = 2.25 mm y=(2n-1) d. (iii) At y = 3 mm, K< D dy 4 Now resultant intensity 1 = 11 + 12 + 2 /1112 cos = 410 + 10 + 2 V45 cos = 510 + 410 cos 3

  3. White light is used in a YDSE with D 1 m and d = 0.9 mm. Light reaching the screen at position y 1 mm is passed through a prism and its spectrum is obtained. Find the missing lines in the visible region of this spectrum. -9 x 104 x 1 x 10-3 m - 900 nm for minima p-(2n-1 )a/2 2AP 1800 = (2n-1) = (2-1) 1800 1800 1800 1800 3 5 7 of these 600 nm and 360 nm lie in the visible range. Hence these will be missing lines in the visible spectrum

  4. A beam of light consisting of wavelengths 6000A and 4500A is used in a YDSE with D 1m and d 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide D 6000 x10-10 x1 d= 10-3 2D 0.45 mm Let nith maxima of 1 and nth maxima of 2 coincide at a position y. then, y n, 1-n, ,-LCM of , and 2 y = LCM of 0.6 mm and 0.45 mm y 1.8 mm Ans. At this point 3rd maxima for 6000 A & 4th maxima for 4500 A coincide

  5. d 10-3 -0.5x10-6 n 1000 is not << 2000 Hence now p-d sin must be used Hence, d sin - --1 000 ysDtan 8275 meter (v) Highest order maxima nmax2000 Hence, n 5000 is not possible.