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Optical Path difference and it's variations
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In this lecture you will get the knowledge of Optical Path Difference due to glass slab.

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

U
Unacademy user
bro what abt the reaction cyclohexene and Cl2 /hu..will it form di halogen mono halogen could u explain with respect to ur first video in the comment
Sachin Rana
2 years ago
It can't form I suppose. The radicals won't be stable.
Stijo joseph
2 years ago
but in the tensor exam 2018 that was the question and the compound formed is 3 chloro cyclohexene was the answer iam asking u that why can't it form 3,6 cyclohexene since we are adding Cl2
Rk
Rupa k
2 years ago
cylcohexene when made to react with chlorine in presence of sunlight, it gives two products 3-Chlorocyclohexene (major) and 1,2-Dichlorohexane (minor). In first case allyl free radical is formed which is more stable due to resonance gives major product and in second case it is addition reaction involves ion mechanism, which is less stable. and coming to 3,6-Dichlorocyclohexene product it is not possible because molecule is symmetrical. you will get only monochloro derivative in case of allyl free radical generation. whether you consider clockwise or anticlockwise it will yield only one monochloro product.
Stijo joseph
2 years ago
thank u rupa..????
bro u r legend.... i was very weak in wave optics u made me to uunderstand all those stuffsso easily..... bro this os a humble request many othem includingme are preparing all for state entrance like kcet mhtcet karnataka vet kerala cet.....so plzzzzz make separate videos in discussion of previous yearcet physics problems
Kartikey Pandey
9 months ago
Sure
Lithesh Shetty
9 months ago
lekin bhai sirf 1.5 mahina bacha hai tho.....please put videos for CET in a great speed ....once again Thank you....
sir , please make a video on diffraction, I am unable to understand concept of diffraction from other teachers. I want read this from you.
please reply for thus message bro.....
is this course completed..?>
Sir diffraction... Plzzz
  1. GEOMETRICAL PATH & OPTICAL PATH



  2. t( -1))


  3. Displacement of fringe on introduction of a glass slab in the path of the light coming out of the slits- On introduction of the thin glass-slab of thickness t and refractive index , the optical path of the ray SP increases by t(u-1). Now the path difference between waves com- ing form S, and S at any point P is Ap S2P - (S,P t (Hu-1)) - (S,P-S,P) t(u 1) ap = d sin -t( -1) yd and Ap D t(u -1) y << D as well. for central bright fringe, Ap 0 - t(u - 1)


  4. ya and Ap- -t( -1) D If y << D as well. for central bright fringe, ap:0 yd t(u-1). The whole fringe pattern gets shifted by the same distance * Notice that this shift is in the direction of the slit before which the glass slab is placed. If the glass slab is placed before the upper slit, the fringe pattern gets shifted upwards and if the glass slab is placed before the lower slit the fringe pattern gets shifted downwards.


  5. Solved Examples In a YDSE with d = 1 mm and D = 1 m, slabs of (t = 1 m, = 3) and (t = 0.5 m, = 2) are introduced in front of upper and lower slit respectively. Find the shift in the fringe pattern. Optical path for light coming from upper slit S, is


  6. S,P +1um (2-1) S2P +0.5 um Similarly optical path for light coming from S, is S2P0.5 um (2-1) S2P 0.5 Lum Path difference : ap-(S2P + 0.5 m)-(SP + 2tm) -(SP-SP)-1.5 m. yd for central bright fringe p-0 1.5um y-1mm x 1m 1.5 mm. The whole pattern is shifted by 1.5 mm upwards.