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