1. ⇒ (MHT CET 2023 13th May Morning Shift )

In a biprism experiment, monochromatic light of wavelength ' $λ$ ' is used. The distance between two coherent sources ' $d$ ' is kept constant. If the distance between slit and eyepiece ' $D$ ' is varied as $D_{1}, D_{2}, D_{3} & D_{4}$ and corresponding measured fringe widths are $Z_{1}, Z_{2}, Z_{3}$ and $Z_{4}$ then

A. $Z_{1} D_{1} = Z_{2} D_{2} = Z_{3} D_{3} = Z_{4} D_{4}$

B. $\frac{Z_{1}}{D_{1}} = \frac{Z_{2}}{D_{2}} = \frac{Z_{3}}{D_{3}} = \frac{Z_{4}}{D_{4}}$

C. $D_{1} \sqrt{Z_{1}} = D_{2} \sqrt{Z_{2}} = D_{3} \sqrt{Z_{3}} = D_{4} \sqrt{Z_{4}}$

D. $Z_{1} \sqrt{D_{1}} = Z_{2} \sqrt{D_{2}} = Z_{3} \sqrt{D_{3}} = Z_{4} \sqrt{D_{4}}$

Correct Option is (B)

$\begin{array}{ll} Fringe width Z = \frac{λ D}{d} \\ ∴ & \frac{Z}{D} = \frac{λ}{d} = constant, as λ and d are constant \\ ∴ & \frac{Z_{1}}{D_{1}} = \frac{Z_{2}}{D_{2}} = \frac{Z_{3}}{D_{3}} = \frac{Z_{4}}{D_{4}} \end{array}$

2. ⇒ (MHT CET 2023 13th May Morning Shift )

$A$ and $B$ are two interfering sources where $A$ is ahead in phase by $54^{\circ}$ relative to B. The observation is taken from point $P$ such that PB $-$ PA = 2.5 $λ$ . Then the phase difference between the waves from A and B reaching point P is (in rad)

A. $3.5 π$

B. $5.3 π$

C. $4.3 π$

D. $5.8 π$

Correct Option is (C)

$\begin{aligned} Total phase difference = ϕ_{1} + ϕ_{2} \\ \begin{aligned} ϕ_{1} & = 54 \times \frac{π}{180} = 0.3 π \\ ϕ_{2} & = \frac{2 π}{λ} \times (PB - PA) \\ = \frac{2 π}{λ} \times 2.5 λ = 5 π \\ ∴ ϕ_{1} & + ϕ_{2} = 5 π + 0.3 π = 5.3 π \end{aligned} \end{aligned}$

3. ⇒ (MHT CET 2023 12th May Evening Shift )

The ratio of intensities of two points on a screen in Young's double slit experiment when waves from the two slits have a path difference of $\frac{λ}{4}$ and $\frac{λ}{6}$ is

$(\cos 90^{\circ} = 0, \cos 60^{\circ} = 0.5)$

A. $2 : 1$

B. $2 : 3$

C. $3 : 4$

D. $3 : 5$

Correct Option is (B)

The intensity at the point due to interference is given as $I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$ ..... (i)

For path difference $\frac{λ}{4}$ , the phase difference is

$ϕ_{1} = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2}$

For path difference $\frac{λ}{6}$ , the phase difference is

$ϕ_{2} = \frac{2 π}{λ} \times \frac{λ}{6} = \frac{π}{3}$

Assuming equal intensity of the interfering waves i.e., $I_{1} = I_{2} = I_{0}$

Equation (i) becomes,

$\begin{aligned} I = I_{0} + I_{0} + 2 I_{0} \cos ϕ \\ I = 2 I_{0} (1 + \cos ϕ) \end{aligned}$

For the given path difference, $I_{1} = 2 I_{0} (1 + \cos \frac{π}{2})$ , and $I_{2} = 2 I_{0} (1 + \cos \frac{π}{3})$

$∴ \frac{I_{1}}{I_{2}} = \frac{1 + \cos \frac{π}{2}}{1 + \cos \frac{π}{3}}$

$\begin{aligned} \frac{I_{1}}{I_{2}} = \frac{1 + 0}{1 + 0.5} \\ ∴ \frac{I_{1}}{I_{2}} & = \frac{1}{1.5} = \frac{2}{3} \end{aligned}$

4. ⇒ (MHT CET 2023 12th May Evening Shift )

In Young's double slit experiment when a glass plate of refractive index 1.44 is introduced in the path of one of the interfering beams, the fringes are displaced by a distance ' $y$ '. If this plate is replaced by another plate of same thickness but of refractive index 1.66, the fringes will be displaced by a distance

A. $\frac{3 y}{2}$

B. $\frac{2 y}{3}$

C. $\frac{5 y}{4}$

D. $\frac{4 y}{5}$

Correct Option is (A)

As a glass plate is used in one of the paths,

$\begin{aligned} y_{1} = \frac{β}{λ} (1.44 - 1) t \\ y_{1} = 0.44 t \times \frac{β}{λ} \end{aligned}$

New displacement is:

$\begin{aligned} y_{2} & = \frac{β}{λ} (1.66 - 1) t \\ y_{2} & = 0.66 t \times \frac{β}{λ} \\ \frac{y_{2}}{y_{1}} & = \frac{0.66}{0.44} \\ ∴ y_{2} & = \frac{3 y}{2} \end{aligned}$

5. ⇒ (MHT CET 2023 12th May Evening Shift )

One of the slits in Young's double slit experiment is covered with a transparent sheet of thickness $2.9 \times 10^{- 3} cm$ . The central fringe shifts to a position originally occupied by the $25^{th}$ bright fringe. If $λ = 5800$ $\overset{o}{A}$ , the refractive index of the sheet is

A. 1.65

B. 1.60

C. 1.55

D. 1.50

Correct Option is (D)

As it is given that a transparent sheet of certain thickness is inserted, we use

$\begin{aligned} (μ - 1) \times t = N λ \\ 25 λ = (μ - 1) t \\ μ - 1 = \frac{25 λ}{t} \end{aligned}$

$∴$ The refractive index of the sheet is:

$\begin{aligned} μ = \frac{25 \times 5800 \times 10^{- 10}}{2.9 \times 10^{- 5}} + 1 \\ μ = 1.50 \end{aligned}$

⇒Wave Optics