Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

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: Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

The document Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev is a part of the Course Physics For JEE.

Q.1. Making use of the spectral response curve for an eye (see Fig. 5.1), find:
 (a) the energy flux corresponding to the luminous flux of 1.0 lm at the wavelengths 0.51 and 0.64 pm;
 (b) the luminous flux corresponding to the wavelength interval from 0.58 to 0.63 tim if the respective energy flux, equal to Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev= 4.5 mW, is uniformly distributed over all wavelengths of the interval.
 The function V (λ) is assumed to be linear in the given spectral interval. 

Ans. (a) The relative spectral response V(λ) shown in Fig. (5.11) of the book is so defined that A/V (λ) is the energy flux of light of wave length λ, needed to produce a unit luminous flux at that wavelength. (A is the conversion factor defined in the book.)

At λ = 0.51 μm, we read from the figure

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

energy flux corresponding to a luminous flux of 1 lumen Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

At

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

and energy flux corresponding to a luminous flux of 1 lumen Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

(b)  Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

since energy is distributed uniformly. Then

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

since V (λ) is assumed to vary linearly in the interval Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev we have

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.2. A point isotropic source emits a luminous flux Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev lm with wavelength λ = 0.59 μm. Find the peak strength values of electric and magnetic fields in the luminous flux at a distance r = = 1.0 m from the source. Make use of the curve illustrated in Fig. 5.1. 

Ans. 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Also Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

 

Q.3. Find the mean illuminance of the irradiated part of an opaque sphere receiving
 (a) a parallel luminous flux resulting in illuminance E at the point of normal incidence;
 (b) light from a point isotropic source located at a distance l = = 100 cm from the centre of the sphere; the radius of the sphere is R = 60 cm and the luminous intensity is I = 36 cd.

Ans. (a) Mean illuminance

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Now, to calculate the total luminous flux incident on the sphere, we note that the illuminance at the point of normal incidence is E 0 . Thus the incident flux is Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRevThus 

Mean illuminance Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

The area irradiated is :Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Thus  Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Substituting we get  Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

 

Q.4. Determine the luminosity of a surface whose luminance depends on direction as L = Lo  cosθ, where θ is the angle between the radiation direction and the normal to the surface. 

Ans. Luminance L is the light energy emitted per unit area of the emitting surface in a given direction per unit solid angle divided by cosθ. Luminosity M is simply energy emitted per unit area.

 Thus

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

where the integration must be in the forward hemisphere of the emitting surface (assuming light is being emitted in only one direction say outward direction of the surface.) But

ThusPhotometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

  

Q.5. A certain luminous surface obeys Lambert's law. Its luminance is equal to L. Find:
 (a) the luminous flux emitted by an element ΔS of this surface into a cone whose axis is normal to the given element and whose aperture angle is equal to θ;
 (b) the luminosity of such a source.

Ans.  For a Lambert source L = Const

The flux emitted into the cone is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

(b) The luminosity is obtained from the previous formula for θ = 900

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.6. An illuminant shaped as a plane horizontal disc S = 100 cm2 in area is suspended over the centre of a round table of radius R = 1.0 m. Its luminance does not depend on direction and is equal to L = 1.6.104  cd/m2. At what height over the table should the illuminant be suspended to provide maximum illuminance at the circumference of the table? How great will that illuminance be? The illuminant is assumed to be a point source. 

Ans. The equivalent luminous intensity in the direction OP  is 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

This is maximum when and the maximum illuminance is        R = h

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

 

Q.7. A point source is suspended at a height h = 1.0 m over the centre of a round table of radius R = 1.0 m. The luminous intensity I of the source depends on direction so that illuminance at all points of the table is the same. Find the function I (θ), where θ is the angle between the radiation direction and the vertical, as well as the luminous flux reaching the table if I (0) = I0 = 100 cd. 

Ans. The illuminance at P is 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

since this is constant at all x, we must have

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 The luminous flux reaching the table is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

 

Q.8. A vertical shaft of light from a projector forms a light spot S = 100 cm2  in area on the ceiling of a round room of radius R = = 2.0 m. The illuminance of the spot is equal to E = 1000 lx. The reflection coefficient of the ceiling is equal to p = 0.80. Find the maximum illuminance of the wall produced by the light reflected from the ceiling. The reflection is assumed to obey Lambert's law. 

Ans. The illuminated area acts as a Lambert source of luminosity M = πL where  

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

The equivalent luminous intensity in the direction making an angle θ from the vertical is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

and the illuminance at the point P is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

This is maximum when

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

orPhotometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Then the maximum illuminance is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

This illuminance is obtained at a distance Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRevfrom the ceiling. Substitution gives the value

0.21 lux

 

Q.9. A luminous dome shaped as a hemisphere rests on a horizontal plane. Its luminosity is uniform. Determine the illuminance at the centre of that plane if its luminance equals L and is independent of direction.

Ans. From the definition of luminance, the energy emitted in the radial direction by an element dS of the surface of the dome is 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

where dA is the area of an element on the plane illuminated by the radial light. Then

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

The illuminance at 0 is then

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.10. A Lambert source has the form of an infinite plane. Its luminance is equal to L. Find the illuminance of an area element oriented parallel to the given source. 

Ans. Consider an element of area dS at point P.
It emits light of flux 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

in the direction of the surface element dA at O.
The total illuminance at O is then

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

But 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Substitution gives Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

 

Q.11. An illuminant shaped as a plane horizontal disc of radius R = 25 cm is suspended over a table at a height h = 75 cm. The illuminance of the table below the centre of the illuminant is equal to E0  = 70 lx. Assuming the source to obey Lambert's law, find its luminosity. 

Ans. Consider an angular element of area 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 Light emitted from this ring is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Now    Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

where dA = an element of area of the table just below the untre of the illuminant Then the illuminance at the element dA will be

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

wherePhotometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRevFinally using luminosity M = πL

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

or  Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.12. A small lamp having the form of a uniformly luminous sphere of radius R = 6.0 cm is suspended at a height h = 3.0 m above the floor. The luminance of the lamp is equal to L = 2.0.104  cd/m2 and is independent of direction. Find the illuminance of the floor directly below the lamp. 

Ans. See the figure below. The light emitted by an element of the illuminant towards the point 0 under consideration is 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

The distance

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Hence, the illuminance at O is

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Substitution gives : Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev 

 

Q.13. Write the law of reflection of a light beam from a mirror in vector form, using the directing unit vectors e and e' of the inci- dent and reflected beams and the unit vector n of the outside normal to the mirror surface. 

Ans. We see from the diagram that because of the law of reflection, the component of the incident unit  

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.14. Demonstrate that a light beam reflected from three mutually perpendicular plane mirrors in succession reverses its direction. 

Ans. We choose the unit vectors perpendicular to the mirror as the x, y,-z axes in space. Then after reflection from the mirror with normal along x axis 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

where Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRevare the basic unit vectors. A fter a second reflection from the 2nd mirror say along y axis.

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Finally after the third reflection

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.15. At what value of the angle of incident θ1 is a shaft of light reflected from the surface of water perpendicular to the refracted shaft?

Ans. Let PQ be the surface of water and n be the R.I. of water. Let AO is the shaft of light with incident angle θand OBand OC are the reflected and refracted light rays at angles 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.16. Two optical media have a plane boundary between them. Supposeθicr is the critical angle of incidence of a beam and θ1  is the angle of incidence at which the refracted beam is perpendicular to the reflected one (the beam is assumed to come from an optically denser medium). Find the relative refractive index of these media if Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Ans. Let two optical mediums of R.I.n1 and n2 respectively be such that n1 >n2 . In the case when angle of incidence is Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev  (Fig.), from the law of refraction 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev              (1)

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

In the case , when the angle of incidence is θ1, from the 

law of refraction at the interface of mediums 1 and 2 .

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

But in accordance with the problem Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

so,        Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev         (2)

Dividing Eqn (1) by (2)

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

or,       Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev(3)

But    Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

So,    Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev        (Using 3)

Thus  Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.17. A light beam falls upon a plane-parallel glass plate d=6.0 cm in thickness. The angle of incidence is θ = 60°. Find the value of deflection of the beam which passed through that plate. 

Ans. From the Fig. the sought lateral shift 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

But from the law of refraction

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

ThusPhotometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

  

Q.18. A man standing on the edge of a swimming pool looks at a stone lying on the bottom. The depth of the swimming pool is equal to h. At what distance from the surface of water is the image of the stone formed if the line of vision makes an angle θ with the normal to the surface? 

Ans. From the Fig.

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Similarly

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

From Eqns (1) and (2)

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev         (3)

From the law of refraction

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev          (A)

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev           (B)

Differentiating Eqn.(A)

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev                  (4)

Using (4) in (3), we get

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev          (5)

 

Q.19. Demonstrate that in a prism with small refracting angle θ the shaft of light deviates through the angle Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev regardless of the angle of incidence, provided that the latter is also small. 

Ans. The figure shows the passage of a monochromatic ray through the given prism, placed in air medium.
From the figure, we have
Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

From the Snell’s law

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

From Eqns (1), (2) and (3), we get

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

So,Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

 

Q.20. A shaft of light passes through a prism with refracting angle θ and refractive index n. Let a be the diffraction angle of the shaft. Demonstrate that if the shaft of light passes through the prism symmetrically,
 (a) the angle a is the least;
 (b) the relationship between the angles α and θ is defined by Eq. (5.1e). 

Ans. (a) In the general case, for the passage of a monochromatic ray through a prism as shown in the figure of the soln. of 5.19, 

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev         (1)

And from the Snell’s law,

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev   (1)

But from the Snell’s law  Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

So, Photometry and Geometrical Optics (Part - 1) - Optics, Irodov JEE Notes | EduRev

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