Optics(12th Grade > Physics ) Questions and Answers
Explanation:-
Answer: Option B. -> 1sini:
B
For glass water interface gμω=sinisinr.....(i)and For water-air interface ωμa=sinrsin90......(ii)
∴gμωxωμa=sini⇒μg=1sini
Question 2. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall. A man walks in front of themirror along a line parallel to the mirror at a distance 2L from it as shown. The greatest distance over which he can see the image of the light source in the mirror is
[IIT-JEE (Screening) 2000]
[IIT-JEE (Screening) 2000]
Explanation:-
Answer: Option C. -> 6 cm:
C
Using refraction formula 1μ2−1R=1μ2v−1u
in given case, medium (1) is glass and (2) is air
So gμa−1R=gμav−1u⇒11.5−1−6=11.5v−1−6
⇒1−1.5−6=1v+1.56⇒0.5v=1v+14
⇒1v=112−14=−212=−16⇒v=6cm.
Explanation:-
Answer: Option A. -> There will be no constant phase difference between the two waves:
A
In conventional light source, light comes from a large number of independent atoms, each atom emitting light for about 10-9 sec i.e. light emitted by an atom is essentially a pulse lasting for only 10-9 sec. Light coming out from two slits will have a fixed phase relationship only for 10-9 sec. Hence any interference pattern formed on the screen would last only for 10-9 sec, and then the pattern will change. The human eye can notice intensity changes which last at least for a tenth of a second and hence we will not be able to see any interference pattern. Instead due to rapid changes in the pattern, we will only observe a uniform intensity over the screen.
Explanation:-
Answer: Option C. -> 7λ2:
C
x=(2n−1)λ2;n=4
=(2×4−1)λ2
=(8−1)λ2
=7λ2
Explanation:-
Answer: Option C. -> 1.20:
C
Refractice index of glass slab =1.5
Velocity of light in glass =2×108ms−1
Velocity of light in liquid =2.50×108ms−1
μα1v⇒μliμg=vgvl⇒μl1.5=2×1082.5×108
⇒μl=1.2