Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The index of refraction of a glass plate is 1.48 at θ1=300C and varies linearly with temperature with a coefficient of 2.5×10−5 0C−1 . The coefficient of linear expansion of the glass is 5×10−5 0C−1 at 300C , the length of the glass plate is 3cm. this plate is placed in front of one of the sits in Young’s double-slit experiment. If the plate is being heated so that its temperature increases at a rate of 5 0C−1min , the light source has wavelength λ=589nm and the glass plate initially is at θ=300C . The number of fringes that shift on the screen in each minute is nearly (use approximation)

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

a

1

b

11

c

110

d

1.1×103

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

μ0=refractive index of glass plateat 30°Cμ=final refractive index of glass plate after heatingt0=initial thickness of glasss plate at 30°Ct=final thickness of glass plate after heatingPath difference =μt−μ0t0=nλ0 Where n is the number of fringes that shift on the screen⇒μ0(1+α1θ)t0(1+α2θ)−μ0t0λ0=nμ0t0(α1+α2)θλ0=nGiven, μ0=1.48,t0=3×10−2mα1=2.5×10−5  0C−1α2=0.5×10−8 0C−1,θ=50min−1λ0=589nm∴n=1.48×3×10−2(3×10−5)×5589×10−9=11
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring