In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5×10−2m towards the slits, the change in fringe width is 3×10−5m If separation between the slits is 10−3m, the wavelength of light used is (in A∘).

β=λDd⇒βαD⇒β1β2=D1D2⇒β1−β2β2=D1−D2D2⇒ΔβΔD=β2D2=λ2d2=λ2=3×10−55×10−2×10−3=6×10−7m=6000A∘

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by $5 \times 10^{- 2} m$ towards the slits, the change in fringe width is $3 \times 10^{- 5} m$ If separation between the slits is $10^{- 3} m$ , the wavelength of light used is (in $A^{\circ}$ ).

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 6000.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$β = \frac{λ D}{d} \Rightarrow β α D \Rightarrow \frac{β_{1}}{β_{2}} = \frac{D_{1}}{D_{2}} \Rightarrow \frac{β_{1} - β_{2}}{β_{2}} = \frac{D_{1} - D_{2}}{D_{2}} \Rightarrow \frac{Δ β}{Δ D} = \frac{β_{2}}{D_{2}} = \frac{λ_{2}}{d_{2}}$

$= λ_{2} = \frac{3 \times 10^{- 5}}{5 \times 10^{- 2}} \times 10^{- 3} = 6 \times 10^{- 7} m = 6000 \overset{\circ}{A}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!