Q 1 :

Average force exerted on a non-reflecting surface at normal incidence is 2.4×10-4N. If 360 W.Cm2 is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is             [2024]

  • 0.02 m2

     

  • 0.2 m2

     

  • 20 m2

     

  • 0.1 m2

     

(1)       

            Radiation force,F=IAcA=FcI

              A=2.4×10-4×3×108360×104=2.4120=0.02m2

 



Q 2 :

An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity 6×108W/m2 falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space =3×108m/s)                 [2024]

  • 36Nm-2

     

  • 18Nm-2

     

  • 6Nm-2

     

  • 2Nm-2

     

(3)       

              Radiation pressure=IV

                                                   =I·μc=6×108×33×108=6N/m2

 



Q 3 :

Two sources of light emit with a power of 200 W. The ratio of the number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be               [2024]

  • 3 : 5

     

  • 1 : 5

     

  • 5 : 3

     

  • 1 : 3

     

(1)      

     Let n1 and n2 are number of photons emitted by sources per second.

            n1×hcλ1=200

           n2×hcλ2=200

           n1n2=λ1λ2=300500=35

 



Q 4 :

Monochromatic light of frequency 6×1014Hz is produced by a laser. The power emitted is 2×10-3W. How many photons per second, on an average, are emitted by the source?               [2024]

(Given h=6.63×10-34 Js)

  • 9×1018

     

  • 6×1015

     

  • 5×1015

     

  • 7×1016

     

(3)         

               Power P=2×10-3W

                Energy E=hν=6.63×10-34×6×1014

                 n=PE=2×10-36.63×10-34×6×1014

                n=5×1015

 



Q 5 :

In the Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by the hydrogen atom when excited to the first excitation level is ___ nm.

(Given hc=1245eVnm, e=1.6×10-19C)                [2024]



(122)            10.2eV=hcλλ=1245 eV-nm10.2eV=122.06nm

 



Q 6 :

If the total energy transferred to a surface in time t is 6.48×109J, then the magnitude of the total momentum delivered to this surface for complete absorption will be   [2024]

  • 2.46×10-3kgm/s

     

  • 2.16×10-3kgm/s

     

  • 1.58×10-3kgm/s

     

  • 4.32×10-3kgm/s

     

(2)

Energy transferred in time 't'=6.48×105J

PowerP=6.48×105t

For complete absorption

Pressure=IC=Energyt.A.C

(Pressure×A)=Energytime×C

Force=Energytime×C

Change in momentum=Energytime×C

So Momentum delivered=EnergyC

=6.48×1053×108=2.16×10-3 kg m/s



Q 7 :

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is          [2025]

  • 1.5×108 Pascals

     

  • 0

     

  • 6×108 Pascals

     

  • 3×108 Pascals

     

(3)

Prad=2Ic

Where I = intensity at surface

c = Speed of light

I=PowerArea=4504πr2=4504π×4=45016π

Prad=2×45016π×3×108=1508π×108

       =5.97×1086×108 Pascals

 



Q 8 :

A small mirror of mass m is suspended by a massless thread of length l. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (c = speed of light in vacuum and g = acceleration due to gravity)          [2025]

  • θ=3E4mcgl

     

  • θ=Emcgl

     

  • θ=E2mcgl

     

  • θ=2Emcgl

     

(4)

Force due to beam assuming complete reflection

F=2Pc=2cdEdt; P is power

So change in momentum of mirror.

m(v0)=Fdt=2cdE=2Ec          ... (i)

Now using work energy theorem

Wg=k

mgl (1cos θ)=012mv2

gl(2 sin2θ2)=v22

as θ is small

gl2 (θ2)2=124E2m2c2      from eq. (i)

glθ2=4E2m2c2  θ=2Emcgl



Q 9 :

The ratio of the power of a light source S1 to that the light source S2 is 2. S1 is emitting 2×1015 photons per second at 600 nm. If the wavelength of the source S2 is 300 nm, then the number of photons per second emitted by S2 is _____ ×1014.         [2025]



(5)

Since power emitted by a source is given as

P=Total energy emittedtime

   =(Energy of photon)×Number of photons (N)t

P1P2=(E1)n1(E2)n2=(hcλ1)n1(hcλ2)n2

P1P2=(λ2λ1)n1n2

Substituting the given values

2=(300600)×2×1015n2

n2=12×1015=5×1014 Photons/sec



Q 10 :

A small object at rest absorbs a light pulse of power 20 mW and duration 300 ns. Assuming speed of light is 3×108m/s, the momentum of the object becomes equal to       [2023]

  • 0.5×10-17 kg m/s

     

  • 2×10-17 kg m/s

     

  • 3×10-17 kg m/s

     

  • 1×10-17 kg m/s

     

(2)

Momentum  p=EnergyC=Power×timeC

p=(20×10-3 W)(300×10-9 s)3×108 m/s=2×10-17 kg-m/s



Q 11 :

A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is            [2023]

  • 12πWm2

     

  • 140πWm2    

     

  • 110πWm2

     

  • 120πWm2

     

(2)

IEF=12×54π×52=140πW/m2



Q 12 :

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24 W. The radius of curvature of hemisphere is 10 cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____ ×10-8 N.                   [2023]



(4)

Force=PdAcosθ

=2ICdAcosθ=2ICπR2=2p04πR2·πR2C

=p02C=242×3×108=4×10-8N



Q 13 :

A photoemissive substance is illuminated with a radiation of wavelength λi so that it releases electrons with de-Broglie wavelength λe. The longest wavelength of radiation that can emit photoelectron is λ0. Expression for de-Broglie wavelength is given by (m = mass of the electron, h = Planck's constant and c = speed of light)         [2025]

  • λe=hλ02mc

     

  • λe=hλi2mc

     

  • λe=h2mc(1λi-1λ0)

     

  • λe=h2mc(1λi-1λ0)

     

(4)

 



Q 14 :

Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is __________. 

(Given: h=6.63×1034J·s and 𝑐=3×108m/s)   [2026]

  • 2×1016

     

  • 5×1016

     

  • 5×1015

     

  • 10×1015

     

(1)

P=nhcλ

6×10-3=n×6.63×10-34×3×108663×10-9

n=2×1016 photons