Failure of Cassical Wave Theory

♦ Learning about : FAILURE OF CLASSICAL WAVE THEORY , Intensity problem ,Frequency Problem , Time delay problem , EINSTEIN’S QUANTUM THEORY OF PHOTO ELECTRIC EFFECT ♦

Intensity problem : According to wave theory intensity of a wave I ∝ E02 and so force on an electron
Fe = eE0, i.e., acceleration and hence, K.E of electron must increase with intensity. However experiments show that Kmax is independent of intensity.

Frequency Problem : According to wave theory photo electric effect should take place for all frequencies provided the intensity of light is sufficient to supply necessary energy to the electron for its emission. However, experiments on photoelectric effect reveal that for every metal there exists a frequency ν0 (depending on the nature of metal) below which no photo electric emission takes place how intense the light be.

Time delay problem : As energy in a wave is spread over the wave front, it will be absorbed by all the electrons to accumulate sufficient energy for their emission from the metal and so according to wave theory there must be a measurable time delay between incidence of light and emission of electrons. However no such time lag has ever been detected experimentally.

Above all clearly shows that photo electric effect and its characteristics can not be accounted on the basis of wave theory of light.

EINSTEIN’S QUANTUM THEORY OF PHOTO ELECTRIC EFFECT :

  • In photoelectric effect, light does not behave as waves but as particles called photons.
  • Photon is a energy packet, i.e., quantum of light energy with no charge and zero rest mass which transfers energy E(= hν) and momentum P(= E/c)
  • Photon electric effect is the result of ‘one to one’ inelastic collision between photon and electron in which photon is completely absorbed. So if an electron in a metal absorbs a photon of energy hν.

When photon is incident on a metal surface, it transfers complete energy to an electron. Some part of energy is used to bring electron at surface and rest of energy is used to increase the K.E. of electron.

Let , Ek =Maximum kinetic energy ,

W = work function of metal ,

hν = energy absorbed by metal

From conservation of energy hν = W + Ek

⇒ Ek = hν – W

since , W = hν0

⇒    Ek = h(ν – ν0 )

⇒   (1/2) mVmax2 = h (ν – ν0) ,

If V0 = stopping potential then ,

e V0 = = h (ν – ν0)

This is known as Einstein photoelectric equation and explains photoelectric effect.

One face of a rectangular glass plate of 8 cm thickness is silvered. An object held at 10 cm in front of ….

Q: One face of a rectangular glass plate of 8 cm thickness is silvered. An object held at 10 cm in front of the unsilvered face forms an image 122 cm behind the silvered face. What is the refractive index of the glass ?

(a) 1.3

(b) 1.4

(c) 1.5

(d) 1.6

Click to See Answer :
Ans: (d)

Sol: Suppose x = Apparent position of silvered face from unsilvered face

Numerical

Distance of object in front of the silvered face = Distance of image behind the silvered face

10 + x = (8-x) + 12

2 x = 8 + 12-10

x = 5 cm

$\displaystyle \mu = \frac{Real \; depth}{Apparent \; depth} $

$\displaystyle \mu = \frac{8}{5} = 1.6 $

 

A fish looking up  through water, sees the outside world through a circular horizon…

Q: A fish looking up  through water, sees the outside world through a circular horizon. If the fish is √7 cm below the surface of water, what is the radius of the circular horizon ?(μ=4/3)

(a) 3 cm

(b) √7 cm

(c) 3 × √7 cm

(d) (3/√7) cm

Click to See Answer :
Ans: (a)

Sol: As we know that , $\displaystyle sinC = \frac{1}{\mu} $

$\displaystyle sinC = \frac{1}{4/3} = \frac{3}{4}$ …(i)

Numerical

Let AB = r = Radius of Circle

$\displaystyle sinC = \frac{AB}{OB} $

$\displaystyle \frac{3}{4} = \frac{r}{OB} $

$\displaystyle OB = \frac{4}{3} r $

In ΔAOB ;

$\displaystyle OA^2 = OB^2 – AB^2$

$\displaystyle 7 = (\frac{4}{3} r)^2 – r^2$

$\displaystyle 7 = \frac{16}{9} r^2 – r^2 = \frac{7}{9} r^2$

r = 3 cm

 

Two identical thin Plano – convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are …

Q: Two identical thin Plano – convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre . The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

(a) – 25 cm

(b) – 50 cm

(c) 50 cm

(d) – 20 cm

Click to See Answer :
Ans: (b)

Sol: From Lens Maker’s Formula ,

$\displaystyle \frac{1}{f} = (\mu -1) (\frac{1}{R_1}-\frac{1}{R_2}) $

Numerical

For First Plano-convex Lens ;

$\displaystyle \frac{1}{f_1} = (1.5 -1) (\frac{1}{\infty}-\frac{1}{-20}) $

$\displaystyle \frac{1}{f_1} = \frac{1}{40} $

For Second Plano-convex Lens ;

$\displaystyle \frac{1}{f_2} = (1.5 -1) (\frac{1}{20}-\frac{1}{\infty}) $

$\displaystyle \frac{1}{f_2} = \frac{1}{40} $

For Concave Lens of Oil ;

$\displaystyle \frac{1}{f_3} = (1.7 -1) (\frac{1}{-20}-\frac{1}{20}) $

$\displaystyle \frac{1}{f_3} = (0.7) (-\frac{2}{-20}) $

$\displaystyle \frac{1}{f_3} = -\frac{7}{100} $

Focal length of the combination is given by ,

$\displaystyle \frac{1}{F} = \frac{1}{f_1} +\frac{1}{f_2} + \frac{1}{f_3}$

$\displaystyle \frac{1}{F} = \frac{1}{40} +\frac{1}{40} – \frac{7}{100}$

f = -50 cm

 

Two identical glass rods S1 and S2 (refractive index=1.5) have one convex end of radius of curvature 10 cm …

Q: Two identical glass rods S1 and S2 (refractive index=1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in Fig with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2. The distance d is

Numerical

(a) 60 cm

(b) 70 cm

(c) 80 cm

(d) 90 cm

Click to See Answer :
Ans: (b)
Sol: For S1 ,

Refraction from denser to rarer medium ,

$\displaystyle \frac{\mu_2}{v} – \frac{\mu_1}{u} = \frac{\mu_2 – \mu_1}{R}$

$\displaystyle \frac{1}{v} – \frac{1.5}{-50} = \frac{1 – 1.5}{-10}$

$\displaystyle \frac{1}{v} = -\frac{1}{20} – \frac{1.5}{50}$

v = 50 cm

For S2 ,

Refraction from rarer to denser medium ,

$\displaystyle \frac{\mu_2}{v} – \frac{\mu_1}{u} = \frac{\mu_2 – \mu_1}{R}$

$\displaystyle \frac{1.5}{\infty} – \frac{1}{-(d-50)} = \frac{1.5 – 1}{10}$

$\displaystyle \frac{1}{(d-50)} = \frac{1}{20}$

d = 70 cm