Q.
In a YDSE experiment, the two slits are covered with a transparent membrane of negligible thickness which allows light to pass through it but does not allow water. A glass slab of thickness t = 0.41 mm and refractive index μg=1.5 is placed in front of one of the slits as shown in the figure. The separation between the slits is d = 0.30 mm. The entire space to the left of the slits is filled with water of refractive index μw=4/3 .A coherent light of intensity I and absolute wavelength λ=5000Ao is being incident on the slits making an angle 30º with horizontal. If screen is placed at a distance D =1m from the slits, then the distance of central maxima from O is n×0.83 cm. Then the value of n is
see full answer
Want to Fund your own JEE / NEET / Foundation preparation ??
Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya
answer is 2.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
If central maxima is formed at point P on the screen, then the path difference due to water medium is cancelled by path difference due to glass slab. For central bright spot, Δx=0⇒S2P+μg−1t−S1P+μwdsin30∘=0⇒S2P−S1P=xdD=μwdsin300−μg−1×t⇒x=−1.66cm
Watch 3-min video & get full concept clarity