The YDSE apparatus is as shown in figure. The condition for point P to be a dark fringe is:

 

($\lambda$=wavelength of light waves)

1. $(l_1-l_3)+(l_2-l_4)=n\lambda$

2. $(l_1-l_2)+(l_3-l_4)=\frac{(2n-1)\lambda }{2}$

3. $(l_1+l_3)+(l_2+l_4)=\frac{(2n-1)\lambda }{2}$

4. $(l_1-l_2)+(l_4-l_3)=\frac{(2n-1)\lambda }{2}$

In a double-slit experiment, two parallel slits are illuminated first by light of wavelength 400 nm and then by the light of unknown wavelength. The fourth-order dark fringe resulting from the known wavelength of light falls in the same place on the screen as the second-order bright fringe from the unknown wavelength. The value of the unknown wavelength of the light is:

1. 900 nm

2. 700 nm

3. 300 nm

4. none of these

In Young's double-slit experiment d/D=10^-4(d=distance between slits, D=distance of screen from the slits). At point P on the screen, the resulting intensity is equal to the intensity due to the individual slit $I_0$. Then, then the distance of point P from the central maximum is $(\lambda =6000 \stackrel{0}{A})$

1. 2 mm

2. 1 mm

3. 0.5 mm

4. 4 mm

In Young's double-slit experiment, the y-coordinates of central maximum and 10th maximum are 2 cm and 5 cm, respectively. When the YDSE apparatus is immersed in a liquid of refractive index 1.5, the corresponding y-coordinates will be:

1. 2 cm, 7.5 cm

2. 3 cm, 6 cm

3. 2 cm, 4 cm

4. 4/3 cm, 103 cm

In Young's double-slit experiment 12 fringes are observed to be formed in a certain segment of the screen when the light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by:

1. 12

2. 18

3. 24

4. 30

In a double-slit experiment, the distance between the slits is d. The screen is at a distance D from the slits. If a bright fringe is formed opposite to a slit on the screen, the order of the fringe is:

1. $\frac{d^2}{2\lambda D}$

2. $\frac{d}{2\lambda D}$

3. $\frac{d^2}{4\lambda D}$

4. $0$

The figure shows two coherent sources $S_1 and S_2$ emitting wavelength $\lambda$. The separation $S_1{S}_2=1.5\lambda and S_1$ is ahead in phase by $\pi /2$ relative to $S_2$. Then the maxima occur in the direction $\theta$ given by ${\sin }^{-1}$ of

(1) $0$

(2) 1/2

(3) -1/6

(4) -5/6

Correct options are

1. (ii), (iii) and (iv)

2. (i), (ii) and (iii)

3. (i), (iii) and (iv)

4. All the above

Let $S_1 and S_2$ be the two slits in Young's double-slit experiment. If central maxima is observed at P and angle $\angle S_{1}P{S}_2=\theta$, then the fringe width for the light of wavelength $\lambda$ will be (assume $\theta$ to be a small angle)

1. $\lambda$/$\theta$

2. $\lambda$$\theta$

3. 2$\lambda$/$\theta$

4. $\lambda$/2$\theta$

In YDSE, coherent monochromatic light having wavelength 600 nm falls on the slits. The first-order bright fringe is at 4.84 mm from central maxima. Determine the wavelength for which the first-order dark fringe will be observed at the same location on the screen. (Take D=3 m)

1. 600 nm

2. 1200 nm

3. 300 nm

4. 900 nm

$\mathrm{Unpolarized} \mathrm{light} \mathrm{of} \mathrm{intensity} 32 {\mathrm{Wm}}^{-2} \mathrm{passes} \mathrm{through} \mathrm{three} \mathrm{polarizers} \mathrm{such} \mathrm{that}$
$\mathrm{transmission} \mathrm{axes} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{and} \mathrm{second} \mathrm{polarizer} \mathrm{makes} \mathrm{an} \mathrm{angle} {30}^0 \mathrm{with} \mathrm{each}$
$\mathrm{other} \mathrm{and} \mathrm{transmission} \mathrm{axis} \mathrm{of} \mathrm{the} \mathrm{last} \mathrm{polarizer} \mathrm{is} \mathrm{crossed} \mathrm{with} \mathrm{that} \mathrm{of} \mathrm{the} \mathrm{first}.\mathrm{The}$
$\mathrm{intensity} \mathrm{the} \mathrm{final} \mathrm{emerging} \mathrm{light} \mathrm{will} \mathrm{be}:$

1. 32 $W{m}^{-2}$

2. 3 $W{m}^{-2}$

3. 8 $W{m}^{-2}$

4. 4 $W{m}^{-2}$