Chapter 10: Wave Optics

The orientation of wavefront and the ray are perpendicular to each other. In the above figure, L1 is the source of parallel rays, forming I2 at the focal length of the lens. The image formed acts as L2 through which the light rays get converged into L2 and the final image I is formed.

We know that the distance between the sun and the earth is very large. We also know that the sun is spherical and can be considered as a point source of light which is present at a very large distance. When the point source is kept at a distance, the wavefront formed will be almost plane.

The frequency of sound waves varies from 20 Hz to 20,000 Hz with a wavelength of 15 m to 15 mm respectively. The diffraction of sound takes place when the size of the slit is comparable to the wavelength of the sound. The wavelength of visible light is 0.4 to 0.7 micron. The diffraction of the light doesn’t take place when the size of the slits is different from the wavelength of the visible light.

Angular separation, φ = 5.8 × 10^–4 rad

The average distance between the two dots = 2.54/300 = 0.85 × 10^-2 cm

At distance z cm, the angle subtended = 0.85 × 10^-2/z

Resolution angle for human = 0.85 × 10^-2/z

Maximum distance between 2 dots = z = 14.5 cm

The above distance is less than the distinct vision, which normal person cannot see.

A polaroid (I) is placed in front of a monochromatic source and polaroid (II) is placed in front of polaroid (I). When the light passes through the polaroid (II), the light is unaffected.

Polaroid (II) is rotated such that there is no passage of light. The rotation is made such that the polaroid (II) is perpendicular to polaroid (I). When the polaroid (II) is rotated there is no passage of light.

When a polaroid (III) is placed between the polaroid (I) and (II) such that the polaroid (III) is not placed perpendicular to the axis, so there is no passage of light.

When the angle of incidence is equal to the Brewster’s angle such that the transmitted light is polarized and reflected light is plane-polarized. Following is the equation:

Such that μ2 < μ1

When the light travels in a medium, the critical angle is given as

Where μ2 < μ1

So, it can be said that |tan iB| > |sin iC| when iB < iC.

5000 Ao = 5000 × 10^-10 m

1/d = 2 sin β/1.22λ

dmin = 1.22 λ/ 2 sin β

λd = 1.22/10 × 10^-10 m

When the 100V light is used, d’min = 1.22 λd/ 2 sin β

d’min = 1.22 × 1.22 × 10^-10/2 sin β

The required ratio = dmin/d’min = 1.22/5000 = 0.244 × 10^-3

The minima will occur when ∆x = S2P – S1P = (2n-1/2)λ

S1P = √D² + (D – x)²

S2P = √D² + (D + x)²

T2P = D + x

T1P = D – x

D = 0.404λ

Amplitude of the wave in perpendicular polarisation

Amplitude of the wave in parallel polarisation

Intensity at the first minima with polariser

∆x = 2d sin θ+(μ-1)L

sin θ0 = -1/16

From central maxima, OP = -D/16

On the positive side,

On the negative side,

The first principal maxima on the positive side is at a distance, which is above point O

The first principal minima on the negative side is at distance, which is below point O