Physics Answer the following in Brief

If a dielectric (non-polar) molecule is placed in an external electric field, a small induced dipole moment is created because the positive charge in each atom is pushed in the direction of the field and negative charge is pushed in the opposite direction as shown in the figure.
Polarization is the amount of induced surface charge per unit area or the surface density of polarization charges appearing at right angles to applied external electric field.

The ratio of the intensities of the two interfering waves is 9 : 1.

Consider a plane wavefront AB of monochromatic light incident obliquely at an angle on a plane mirror MN as shown in the figure.

MN : plane mirror, RA and QC : Incident rays, AP and CS : Normals to MN, AB : Incident wavefront, CE : Reflected wavefront, i : Angle of incidence, r : Angle of reflection
The wave front AB touches the reflecting surface MN at A at time t = 0.
Let v be the speed of light in the medium. lf T is the time taken by the incident wavefront to travel from B to C.
Then BC = v T.
During this time, secondary wave originating from A covers the same distance, so that the secondary spherical wavelet has a radius vT at time T. Drow a tangent CE to the wavelet.
From figure,
∠RAP = ∠i = angle of incidence
∠PAE = ∠r = angle of redlection
In ΔABC and Δ AEC
AC is common side
AE = BE
∠ABC = ∠AEC
= 90°
∴ ΔABC and ΔAEC are congruent.
∴ ∠ACE = ∠BAC
= ∠i ....(i)
Also as AE is perpendicular to CE and AP is perpendicular to AC
∠ACE = ∠PAE
= ∠r ...(ii)
∴ From equations (i) and (ii),
∠ i = ∠ r
As a result, the angle of incidence and the angle of reflection are identical. The first rule of reflection is this.
Additionally, it is clear from the figure that incident and reflected rays both reside in the same plane but on opposing sides of the normal. The second low of reflection is at hand.
Thus, the laws of reflection of light ore deduced.

Given,
Refractive index of glass with respect to air, \(\mu_g=1.5\)
Refractive index of water with respect to air, \(\mu_w=\frac{4}{3}\)
Let \(v_g\) be speed of light in glass and
Let \(v_w\) be speed of light in water
The \(\left|v_g-v_w\right|=0.25 \times 10^8\)
\(\mu_g=\frac{\text { speed of light in air }}{\text { speed of light in glass }}\)
So, \(v_g=\frac{v_a}{\mu_g}\) and \(v_w=\frac{v_a}{\mu_w}\)
\(\Rightarrow\left|v_g \sim v_w\right|=0.25 \times 10^8\)
\(\Rightarrow\left|\frac{v_a}{\mu_g}-\frac{v_a}{\mu_w}\right|=0.25 \times 10^8\)
\(\Rightarrow v_a\left|\frac{1}{\mu_g}-\frac{1}{\mu_w}\right|=0.25 \times 10^8\)
\(\Rightarrow v_a\left|\frac{1}{1.5}-\frac{3}{4}\right|=0.25 \times 10^8\)
\(\Rightarrow v_a \times 0.0833=0.25 \times 10^8\)
\(\therefore\) Speed of light in air \(v_a=\frac{0.25 \times 10^8}{0.833}=3 \times 10^8 m / s\)

\(\lambda=5200Å=5.2 \times 10^{-7} m\)
\(X_1-X_2=1.3 mm =1.3 \times 10^{-3} m\)
\(D_1-D_2=50 cm =0.5 m\)
\(X_1-X_2=\frac{\lambda D_1}{d}-\frac{\lambda D_2}{d}\)
\(=\frac{\lambda}{d}\left(D_1-D_2\right)\)
\(\therefore d=\frac{\lambda\left(D_1-D_2\right)}{X_1-X_2}\)
\(=\frac{5.2 \times 10^{-7} \times 0.5}{1.3 \times 10^{-3}}\)
\(=2 \times 10^{-4} m\)
= 0.2 mm

\(\lambda=5200Å=5.2 \times 10^{-7} m\)
\(X_1-X_2=1.3 mm =1.3 \times 10^{-3} m\)
\(D_1-D_2=50 cm =0.5 m\)
\(X_1-X_2=\frac{\lambda D_1}{d}-\frac{\lambda D_2}{d}\)
\(=\frac{\lambda}{d}\left(D_1-D_2\right)\)
\(\therefore d=\frac{\lambda\left(D_1-D_2\right)}{X_1-X_2}\)
\(=\frac{5.2 \times 10^{-7} \times 0.5}{1.3 \times 10^{-3}}\)
\(=2 \times 10^{-4} m\)
= 0.2 mm

\({ }^a \mu_w=\frac{4}{3}\)
\({ }^a \mu_d=2.42\)
\({ }^a \mu_w=\frac{4}{3}=\frac{V_a}{V_w}\)
\(\therefore V_w=\frac{3 \times 10^8}{\frac{4}{3}}=3 \times 10^8 \times \frac{3}{4}=2.25 \times 10^8 m / c\)
\({ }^a \mu_d=2.42=\frac{V_a}{V_d}=\frac{3 \times 10^8}{V_d}\)
\(V_d=\frac{3 \times 10^8}{2.42}=1.23 \times 10^8 m / s\)

Given: i = 65°, CD = 2AB
To find: Refractive index (μ)
formulae:
\(\frac{\cos i}{\cos r}=\frac{ AB }{ CD } \ldots \ldots \ldots . .(1)\)
\(\mu=\frac{\sin i}{\sin r}\)........(2)
From formula 1
\(\frac{\cos 65^{\circ}}{\cos r}=\frac{ AB }{2 AB }\)
\(\frac{0.4226}{\cos r}=\frac{1}{2}\)
\(\cos r=32^{\circ} 16 \prime\)
from formula 2
\(\mu=\frac{\sin 65^{\circ}}{\sin \left(32^{\circ} 16 \prime\right)}\)
\(\mu=1.697\)
The refractive index for the denser medium is 1.697.

\(\frac{I_{\max }}{I_{\min }}=\left(\frac{\frac{A_1}{A_2}+1}{\frac{A_1}{A_2}-1}\right)^2\)
\(\operatorname{Let} \frac{A_1}{A_2}=r\)
\(\therefore \frac{I_{\max }}{I_{\min }}=\left(\frac{r+1}{r-1}\right)^2\)
\(\therefore \frac{36}{16}=\left(\frac{r+1}{r-1}\right)^2\)
\(\therefore \frac{6}{4}=\left(\frac{r+1}{r-1}\right)\)
\(\therefore r=\frac{5}{1}\)
\(\frac{A_1}{A_2}=\frac{5}{1}\)
Ratio of width \(=\left(\frac{A_1}{A_2}\right)^2=\frac{25}{1}\)

Huygen’s wave theory :
Reflection at a plane surface : Consider a plane reflecting surface XY. Let AB be a plane wavefront of light incident obliquely on XY. When the incident wavefront touches XY at A, a secondary wavelet starts, spreading from A according to Huygens’ principle. Let the ray at B reach XY at D after a time t. If v is the speed of light in air then BD = vt. During this time t, the secondary wavelet from A spreads over a hemisphere of radius vt. with centre at A. Let CD be a tangent to this hemisphere. Then AC = BD. C and D are in the same phase. If we consider all the points between A and D, then CD will be tangential to all the secondary wavelets originating from these points at the end of t seconds. Hence CD is the reflected wavefront.

Draw AN normal to XY. Then
\(\angle P A N=i\), the angle of incidence, and \(\angle N A C=r\) the angle of reflection
In triangles BAD and CDA
AC = BD = vt; AD is common, and \(\angle A B D=\angle A C D=90^{\circ}\)
because the rays are normal to wavefronts.
\(\therefore\) Triangles BAD and CDA are congruent
\(\therefore \angle D A C=\angle B D A, 90^{\circ}-r=90^{\circ}-i\)
OR \(\therefore i=r\)
i.e , angle of incidence is equal to angle of reflection . 

Given : \(\mu_8=1.5 ; n =4 \times 10^{14} Hz ; c =3 \times 10^8 m / s\)
The wavelength of light incident on glass from air is
\(x _{ a }=\frac{ c }{ n }=\frac{3 \times 10^8}{4 \times 10^{14}}=7.5 \times 10^{-7} m =7500 \times 10^{-10}=7500 A\)
Now, the velocity of light in glass is given from its refractive index as
\(\mu_g=\frac{c}{v_g}\)
We also know that velocity is product of frequency and wavelength.
\(\therefore \mu_g=\frac{c}{v_g}=\frac{n \lambda_\sigma}{n \lambda_g}=\frac{\lambda_\sigma}{\lambda_g}\)
\(\therefore \lambda_g=\frac{\lambda_\sigma}{\lambda_g}=\frac{7500}{1.5}=\)5000Å
Therefore, the difference in wavelength is
\(\lambda_\sigma-\lambda_g=7500-5000=\)2500Å)

\(6469 \AA\)

coming soon…

Given:
\({ }_a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}\)
Find: \({ }_w \mu_g\)
Formula: \({ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}\)
solution: \(a_{\mu-} g=\frac{C_a}{C_g}\) and \({ }_a \mu_w=\frac{C_a}{C_w}\)
\({ }_w \mu_g=\frac{C_w}{C_g}\)
From Formula
\({ }_a \mu_g=\frac{\frac{3}{2}}{\frac{4}{3}}\)
\({ }_w \mu_g=1.12\)
The refractive index of glass w.r.t. water is 1.12.