Wave Optics — JEE Main practice

1. When a transparent sheet of thickness $t$ and refractive index $n$ is placed in front of one slit, the light travels more slowly inside the sheet than it would in air.

The additional optical path difference introduced is

$$(n-1)t.$$

2. In Young's double slit experiment, the geometric path difference for a point at a distance $y$ from the central axis is

$$\frac{dy}{D},$$

where $d$ is the slit separation and $D$ is the distance of the screen from the slits.

3. The shifted central fringe is the position where the additional optical path difference is exactly balanced by the geometric path difference.

Therefore,

$$\begin{aligned} \frac{dy}{D} &=(n-1)t. \end{aligned}$$

4. Rearranging,

$$\begin{aligned} t &=\frac{yd}{D(n-1)}. \end{aligned}$$

5. Substituting the given values,

$$y=0.2\ \text{cm},\quad d=0.1\ \text{cm},\quad D=50\ \text{cm},\quad n=1.5,$$ $$\begin{aligned} t &=\frac{0.2\times0.1}{50(1.5-1)} \\ &=\frac{0.02}{25} \\ &=8\times10^{-4}\ \text{cm}. \end{aligned}$$

6. Hence, the required thickness of the sheet is

$$\boxed{8\times10^{-4}\ \text{cm}},$$

so the correct option is B.

1. In Young's double slit experiment, the position of the $n^{\text{th}}$ bright fringe is

   $$y_n=\frac{n\lambda D}{d},$$

   where:

   • $\lambda$ is the wavelength of light,
   • $D$ is the distance between the slits and the screen,
   • $d$ is the separation between the slits.

2. The linear fringe width is

   $$\beta=y_{n+1}-y_n=\frac{\lambda D}{d}.$$

3. The angular fringe width is the angle subtended by one fringe at the slits. For small angles,

   $$\theta_\beta=\frac{\beta}{D}.$$

   Substituting the value of $\beta$,

   $$\theta_\beta=\frac{\frac{\lambda D}{d}}{D} =\frac{\lambda}{d}.$$

4. From this expression,

   $$\theta_\beta\propto\lambda \qquad\text{and}\qquad \theta_\beta\propto\frac{1}{d}.$$

   Hence, the angular fringe width is independent of the screen distance $D$.

   Therefore, Statement I is false.

5. Since

   $$\theta_\beta=\frac{\lambda}{d},$$

   increasing the wavelength increases the angular separation of the fringes.

   Therefore, Statement II is true.

6. Thus, Statement I is false and Statement II is true.

Hence, Option C is correct.

1. In a single-slit diffraction experiment, the linear width of the central maximum is

   $$W=\frac{2\lambda D}{a},$$

   where:

   • $\lambda$ is the wavelength of light,
   • $D$ is the distance between the slit and the screen,
   • $a$ is the slit width.

2. From the formula,

   $$W\propto\lambda$$

   when the slit width is constant.

   Therefore, increasing the wavelength increases the width of the central maximum.

   • Statement A is true.
   • Statement B is false.

3. Also,

   $$W\propto\frac{1}{a}$$

   when the wavelength is constant.

   Thus, decreasing the slit width increases the width of the central maximum.

   • Statement C is true.
   • Statement D is false.

4. For a fixed slit width, the total transmitted power remains the same. If the wavelength decreases, the diffraction pattern becomes narrower, so the same energy is concentrated into a smaller region.

   Hence, the peak brightness of the central maximum increases.

   • Statement E is true.

5. Therefore, the correct statements are

   $$\boxed{\text{A, C, and E}.}$$

Hence, Option B is correct.