QBManager

WAVE OPTICS

283180 Interference fringes are obtained in a Young's double slit experiment using beam of light consisting two wavelengths $500 nm$ and 600 $nm$ . Bright fringes due to both wavelengths coincide at $2.5 mm$ from the central maximum. If the separation between the slits is $3 mm$ , then the distance between the screen and plane of the slits is

1

1.2 m

2

2.8 m

3

2.5 m

4

3.2 m

Explanation:

: Given that,
$λ_{1} = 500 \times 10^{- 9} m = 5 \times 10^{- 7} m$
$λ_{2} = 600 \times 10^{- 9} m = 6 \times 10^{- 7} m$
$d = 3 mm = 3 \times 10^{- 3} m$
original image
We know that,
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$y_{2} = \frac{n_{2} λ_{2} D}{d}$
$y_{1} = y_{2}$
$n_{1} λ_{1} = n_{2} λ_{2}$
$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{n_{1}}{n_{2}} = \frac{6 \times 10^{- 7}}{5 \times 10^{- 7}} = \frac{6}{5}$
$n_{1} = 6, n_{2} = 5$
From equation (i), we get-
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$2.5 \times 10^{- 3} = \frac{6 \times 5 \times 10^{- 7} \times D}{3 \times 10^{- 3}} (y = 2.5 \times 10^{- 3} m)$
$D = 2.5 m$

TS- EAMCET-04.05.2018

WAVE OPTICS

283181 In an experiment, light passing through two slits separated by a distance of $0.3 mm$ is projected on to a screen placed at $1 m$ from the plane of slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 mm$ . The wavelength of light in $nm$ is

1 450

2 495

3 530

4 570

Explanation:

: Given, $d = 0.3 mm = 0.3 \times 10^{- 3} m, D = 1 m$
original image
Fringe width-
$β = 1.9 mm = 1.9 \times 10^{- 3} m$
Now, we know-
$β = \frac{λ D}{d}$
$λ = \frac{β d}{D} = \frac{1.9 \times 10^{- 3} \times 0.3 \times 10^{- 3}}{1}$
$λ = 0.57 \times 10^{- 6}$
$λ = 570 \times 10^{- 9} m$
$λ = 570 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283182 Calculated the minimum thickness of a soap film $(μ = 1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 nm$ .

1

113 nm

2

100 nm

3

200 nm

4

226 nm

Explanation:

: Given that,
Refractive index $(μ) = 1.33$
$λ = 532 nm = 532 \times 10^{- 9} m$
We know that,
$2 μ t = (n - \frac{1}{2}) λ,$
For minimum thickness $(n) = 1$
$2 μ t = (1 - \frac{1}{2}) λ$
$2 \times 1.33 \times t = \frac{1}{2} λ$
$t = \frac{λ}{2 \times 2 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{4 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{5.32}$
$t = 100 \times 10^{- 9} m$
$t = 100 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283183 In Young's double slit experiment, the intensity of light coming from one of the slits is double the intensity from the other slit.
The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

1 34

2 40

3 25

4 38

Explanation:

: Given that,
$I_{1} = 2 I_{2}$
We know that,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{\frac{2 I_{2}}{I_{2}}} + 1}{\sqrt{\frac{2 I_{2}}{I_{2}}} - 1})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{1.414 + 1}{1.414 - 1})}^{2}$
$\frac{I_{max}}{I_{min}} ◻ 34 : 1$ Hence, the ratio of the maximum to the minimum intensity in the interference fringe pattern observed is
34

Manipal UGET-2018

WAVE OPTICS

283185 Two coherent sources of equal intensities produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by $50 %$ by reducing its width, then the intensity of light at the same will be units.

1 90.35

2 89.86

3 67.35

4 72.85

Explanation:

: Given,
$I_{max} = 100$
$I_{max} = I_{o} + I_{o} + 2 \sqrt{I_{0} I_{0}}$
$4 I_{0} = 100$
$I_{0} = 25$
If the intensity of one of the source is reduced by $50 %$ then, $I_{1} = 25$
$I_{2} = 25 - 25 \times \frac{50}{100}$
$I_{2} = 12.5$
$I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cdot \cos θ$
$I_{R} = 25 + 12.5 + 2 \sqrt{25 \times 12.5} \cos 0^{\circ}$
$I_{R} = 37.5 + 2 (17.67)$
$I_{R} = 37.5 + 35.35$
$I_{R} = 72.85$

AP EAMCET-25.04.2018

WAVE OPTICS

283180 Interference fringes are obtained in a Young's double slit experiment using beam of light consisting two wavelengths $500 nm$ and 600 $nm$ . Bright fringes due to both wavelengths coincide at $2.5 mm$ from the central maximum. If the separation between the slits is $3 mm$ , then the distance between the screen and plane of the slits is

1

1.2 m

2

2.8 m

3

2.5 m

4

3.2 m

Explanation:

: Given that,
$λ_{1} = 500 \times 10^{- 9} m = 5 \times 10^{- 7} m$
$λ_{2} = 600 \times 10^{- 9} m = 6 \times 10^{- 7} m$
$d = 3 mm = 3 \times 10^{- 3} m$
original image
We know that,
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$y_{2} = \frac{n_{2} λ_{2} D}{d}$
$y_{1} = y_{2}$
$n_{1} λ_{1} = n_{2} λ_{2}$
$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{n_{1}}{n_{2}} = \frac{6 \times 10^{- 7}}{5 \times 10^{- 7}} = \frac{6}{5}$
$n_{1} = 6, n_{2} = 5$
From equation (i), we get-
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$2.5 \times 10^{- 3} = \frac{6 \times 5 \times 10^{- 7} \times D}{3 \times 10^{- 3}} (y = 2.5 \times 10^{- 3} m)$
$D = 2.5 m$

TS- EAMCET-04.05.2018

WAVE OPTICS

283181 In an experiment, light passing through two slits separated by a distance of $0.3 mm$ is projected on to a screen placed at $1 m$ from the plane of slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 mm$ . The wavelength of light in $nm$ is

1 450

2 495

3 530

4 570

Explanation:

: Given, $d = 0.3 mm = 0.3 \times 10^{- 3} m, D = 1 m$
original image
Fringe width-
$β = 1.9 mm = 1.9 \times 10^{- 3} m$
Now, we know-
$β = \frac{λ D}{d}$
$λ = \frac{β d}{D} = \frac{1.9 \times 10^{- 3} \times 0.3 \times 10^{- 3}}{1}$
$λ = 0.57 \times 10^{- 6}$
$λ = 570 \times 10^{- 9} m$
$λ = 570 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283182 Calculated the minimum thickness of a soap film $(μ = 1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 nm$ .

1

113 nm

2

100 nm

3

200 nm

4

226 nm

Explanation:

: Given that,
Refractive index $(μ) = 1.33$
$λ = 532 nm = 532 \times 10^{- 9} m$
We know that,
$2 μ t = (n - \frac{1}{2}) λ,$
For minimum thickness $(n) = 1$
$2 μ t = (1 - \frac{1}{2}) λ$
$2 \times 1.33 \times t = \frac{1}{2} λ$
$t = \frac{λ}{2 \times 2 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{4 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{5.32}$
$t = 100 \times 10^{- 9} m$
$t = 100 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283183 In Young's double slit experiment, the intensity of light coming from one of the slits is double the intensity from the other slit.
The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

1 34

2 40

3 25

4 38

Explanation:

: Given that,
$I_{1} = 2 I_{2}$
We know that,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{\frac{2 I_{2}}{I_{2}}} + 1}{\sqrt{\frac{2 I_{2}}{I_{2}}} - 1})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{1.414 + 1}{1.414 - 1})}^{2}$
$\frac{I_{max}}{I_{min}} ◻ 34 : 1$ Hence, the ratio of the maximum to the minimum intensity in the interference fringe pattern observed is
34

Manipal UGET-2018

WAVE OPTICS

283185 Two coherent sources of equal intensities produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by $50 %$ by reducing its width, then the intensity of light at the same will be units.

1 90.35

2 89.86

3 67.35

4 72.85

Explanation:

: Given,
$I_{max} = 100$
$I_{max} = I_{o} + I_{o} + 2 \sqrt{I_{0} I_{0}}$
$4 I_{0} = 100$
$I_{0} = 25$
If the intensity of one of the source is reduced by $50 %$ then, $I_{1} = 25$
$I_{2} = 25 - 25 \times \frac{50}{100}$
$I_{2} = 12.5$
$I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cdot \cos θ$
$I_{R} = 25 + 12.5 + 2 \sqrt{25 \times 12.5} \cos 0^{\circ}$
$I_{R} = 37.5 + 2 (17.67)$
$I_{R} = 37.5 + 35.35$
$I_{R} = 72.85$

AP EAMCET-25.04.2018

WAVE OPTICS

283180 Interference fringes are obtained in a Young's double slit experiment using beam of light consisting two wavelengths $500 nm$ and 600 $nm$ . Bright fringes due to both wavelengths coincide at $2.5 mm$ from the central maximum. If the separation between the slits is $3 mm$ , then the distance between the screen and plane of the slits is

1

1.2 m

2

2.8 m

3

2.5 m

4

3.2 m

Explanation:

: Given that,
$λ_{1} = 500 \times 10^{- 9} m = 5 \times 10^{- 7} m$
$λ_{2} = 600 \times 10^{- 9} m = 6 \times 10^{- 7} m$
$d = 3 mm = 3 \times 10^{- 3} m$
original image
We know that,
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$y_{2} = \frac{n_{2} λ_{2} D}{d}$
$y_{1} = y_{2}$
$n_{1} λ_{1} = n_{2} λ_{2}$
$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{n_{1}}{n_{2}} = \frac{6 \times 10^{- 7}}{5 \times 10^{- 7}} = \frac{6}{5}$
$n_{1} = 6, n_{2} = 5$
From equation (i), we get-
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$2.5 \times 10^{- 3} = \frac{6 \times 5 \times 10^{- 7} \times D}{3 \times 10^{- 3}} (y = 2.5 \times 10^{- 3} m)$
$D = 2.5 m$

TS- EAMCET-04.05.2018

WAVE OPTICS

283181 In an experiment, light passing through two slits separated by a distance of $0.3 mm$ is projected on to a screen placed at $1 m$ from the plane of slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 mm$ . The wavelength of light in $nm$ is

1 450

2 495

3 530

4 570

Explanation:

: Given, $d = 0.3 mm = 0.3 \times 10^{- 3} m, D = 1 m$
original image
Fringe width-
$β = 1.9 mm = 1.9 \times 10^{- 3} m$
Now, we know-
$β = \frac{λ D}{d}$
$λ = \frac{β d}{D} = \frac{1.9 \times 10^{- 3} \times 0.3 \times 10^{- 3}}{1}$
$λ = 0.57 \times 10^{- 6}$
$λ = 570 \times 10^{- 9} m$
$λ = 570 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283182 Calculated the minimum thickness of a soap film $(μ = 1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 nm$ .

1

113 nm

2

100 nm

3

200 nm

4

226 nm

Explanation:

: Given that,
Refractive index $(μ) = 1.33$
$λ = 532 nm = 532 \times 10^{- 9} m$
We know that,
$2 μ t = (n - \frac{1}{2}) λ,$
For minimum thickness $(n) = 1$
$2 μ t = (1 - \frac{1}{2}) λ$
$2 \times 1.33 \times t = \frac{1}{2} λ$
$t = \frac{λ}{2 \times 2 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{4 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{5.32}$
$t = 100 \times 10^{- 9} m$
$t = 100 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283183 In Young's double slit experiment, the intensity of light coming from one of the slits is double the intensity from the other slit.
The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

1 34

2 40

3 25

4 38

Explanation:

: Given that,
$I_{1} = 2 I_{2}$
We know that,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{\frac{2 I_{2}}{I_{2}}} + 1}{\sqrt{\frac{2 I_{2}}{I_{2}}} - 1})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{1.414 + 1}{1.414 - 1})}^{2}$
$\frac{I_{max}}{I_{min}} ◻ 34 : 1$ Hence, the ratio of the maximum to the minimum intensity in the interference fringe pattern observed is
34

Manipal UGET-2018

WAVE OPTICS

283185 Two coherent sources of equal intensities produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by $50 %$ by reducing its width, then the intensity of light at the same will be units.

1 90.35

2 89.86

3 67.35

4 72.85

Explanation:

: Given,
$I_{max} = 100$
$I_{max} = I_{o} + I_{o} + 2 \sqrt{I_{0} I_{0}}$
$4 I_{0} = 100$
$I_{0} = 25$
If the intensity of one of the source is reduced by $50 %$ then, $I_{1} = 25$
$I_{2} = 25 - 25 \times \frac{50}{100}$
$I_{2} = 12.5$
$I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cdot \cos θ$
$I_{R} = 25 + 12.5 + 2 \sqrt{25 \times 12.5} \cos 0^{\circ}$
$I_{R} = 37.5 + 2 (17.67)$
$I_{R} = 37.5 + 35.35$
$I_{R} = 72.85$

AP EAMCET-25.04.2018

WAVE OPTICS

283180 Interference fringes are obtained in a Young's double slit experiment using beam of light consisting two wavelengths $500 nm$ and 600 $nm$ . Bright fringes due to both wavelengths coincide at $2.5 mm$ from the central maximum. If the separation between the slits is $3 mm$ , then the distance between the screen and plane of the slits is

1

1.2 m

2

2.8 m

3

2.5 m

4

3.2 m

Explanation:

: Given that,
$λ_{1} = 500 \times 10^{- 9} m = 5 \times 10^{- 7} m$
$λ_{2} = 600 \times 10^{- 9} m = 6 \times 10^{- 7} m$
$d = 3 mm = 3 \times 10^{- 3} m$
original image
We know that,
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$y_{2} = \frac{n_{2} λ_{2} D}{d}$
$y_{1} = y_{2}$
$n_{1} λ_{1} = n_{2} λ_{2}$
$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{n_{1}}{n_{2}} = \frac{6 \times 10^{- 7}}{5 \times 10^{- 7}} = \frac{6}{5}$
$n_{1} = 6, n_{2} = 5$
From equation (i), we get-
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$2.5 \times 10^{- 3} = \frac{6 \times 5 \times 10^{- 7} \times D}{3 \times 10^{- 3}} (y = 2.5 \times 10^{- 3} m)$
$D = 2.5 m$

TS- EAMCET-04.05.2018

WAVE OPTICS

283181 In an experiment, light passing through two slits separated by a distance of $0.3 mm$ is projected on to a screen placed at $1 m$ from the plane of slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 mm$ . The wavelength of light in $nm$ is

1 450

2 495

3 530

4 570

Explanation:

: Given, $d = 0.3 mm = 0.3 \times 10^{- 3} m, D = 1 m$
original image
Fringe width-
$β = 1.9 mm = 1.9 \times 10^{- 3} m$
Now, we know-
$β = \frac{λ D}{d}$
$λ = \frac{β d}{D} = \frac{1.9 \times 10^{- 3} \times 0.3 \times 10^{- 3}}{1}$
$λ = 0.57 \times 10^{- 6}$
$λ = 570 \times 10^{- 9} m$
$λ = 570 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283182 Calculated the minimum thickness of a soap film $(μ = 1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 nm$ .

1

113 nm

2

100 nm

3

200 nm

4

226 nm

Explanation:

: Given that,
Refractive index $(μ) = 1.33$
$λ = 532 nm = 532 \times 10^{- 9} m$
We know that,
$2 μ t = (n - \frac{1}{2}) λ,$
For minimum thickness $(n) = 1$
$2 μ t = (1 - \frac{1}{2}) λ$
$2 \times 1.33 \times t = \frac{1}{2} λ$
$t = \frac{λ}{2 \times 2 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{4 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{5.32}$
$t = 100 \times 10^{- 9} m$
$t = 100 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283183 In Young's double slit experiment, the intensity of light coming from one of the slits is double the intensity from the other slit.
The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

1 34

2 40

3 25

4 38

Explanation:

: Given that,
$I_{1} = 2 I_{2}$
We know that,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{\frac{2 I_{2}}{I_{2}}} + 1}{\sqrt{\frac{2 I_{2}}{I_{2}}} - 1})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{1.414 + 1}{1.414 - 1})}^{2}$
$\frac{I_{max}}{I_{min}} ◻ 34 : 1$ Hence, the ratio of the maximum to the minimum intensity in the interference fringe pattern observed is
34

Manipal UGET-2018

WAVE OPTICS

283185 Two coherent sources of equal intensities produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by $50 %$ by reducing its width, then the intensity of light at the same will be units.

1 90.35

2 89.86

3 67.35

4 72.85

Explanation:

: Given,
$I_{max} = 100$
$I_{max} = I_{o} + I_{o} + 2 \sqrt{I_{0} I_{0}}$
$4 I_{0} = 100$
$I_{0} = 25$
If the intensity of one of the source is reduced by $50 %$ then, $I_{1} = 25$
$I_{2} = 25 - 25 \times \frac{50}{100}$
$I_{2} = 12.5$
$I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cdot \cos θ$
$I_{R} = 25 + 12.5 + 2 \sqrt{25 \times 12.5} \cos 0^{\circ}$
$I_{R} = 37.5 + 2 (17.67)$
$I_{R} = 37.5 + 35.35$
$I_{R} = 72.85$

AP EAMCET-25.04.2018

WAVE OPTICS

283180 Interference fringes are obtained in a Young's double slit experiment using beam of light consisting two wavelengths $500 nm$ and 600 $nm$ . Bright fringes due to both wavelengths coincide at $2.5 mm$ from the central maximum. If the separation between the slits is $3 mm$ , then the distance between the screen and plane of the slits is

1

1.2 m

2

2.8 m

3

2.5 m

4

3.2 m

Explanation:

: Given that,
$λ_{1} = 500 \times 10^{- 9} m = 5 \times 10^{- 7} m$
$λ_{2} = 600 \times 10^{- 9} m = 6 \times 10^{- 7} m$
$d = 3 mm = 3 \times 10^{- 3} m$
original image
We know that,
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$y_{2} = \frac{n_{2} λ_{2} D}{d}$
$y_{1} = y_{2}$
$n_{1} λ_{1} = n_{2} λ_{2}$
$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}}$
$\frac{n_{1}}{n_{2}} = \frac{6 \times 10^{- 7}}{5 \times 10^{- 7}} = \frac{6}{5}$
$n_{1} = 6, n_{2} = 5$
From equation (i), we get-
$y_{1} = \frac{n_{1} λ_{1} D}{d}$
$2.5 \times 10^{- 3} = \frac{6 \times 5 \times 10^{- 7} \times D}{3 \times 10^{- 3}} (y = 2.5 \times 10^{- 3} m)$
$D = 2.5 m$

TS- EAMCET-04.05.2018

WAVE OPTICS

283181 In an experiment, light passing through two slits separated by a distance of $0.3 mm$ is projected on to a screen placed at $1 m$ from the plane of slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 mm$ . The wavelength of light in $nm$ is

1 450

2 495

3 530

4 570

Explanation:

: Given, $d = 0.3 mm = 0.3 \times 10^{- 3} m, D = 1 m$
original image
Fringe width-
$β = 1.9 mm = 1.9 \times 10^{- 3} m$
Now, we know-
$β = \frac{λ D}{d}$
$λ = \frac{β d}{D} = \frac{1.9 \times 10^{- 3} \times 0.3 \times 10^{- 3}}{1}$
$λ = 0.57 \times 10^{- 6}$
$λ = 570 \times 10^{- 9} m$
$λ = 570 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283182 Calculated the minimum thickness of a soap film $(μ = 1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 nm$ .

1

113 nm

2

100 nm

3

200 nm

4

226 nm

Explanation:

: Given that,
Refractive index $(μ) = 1.33$
$λ = 532 nm = 532 \times 10^{- 9} m$
We know that,
$2 μ t = (n - \frac{1}{2}) λ,$
For minimum thickness $(n) = 1$
$2 μ t = (1 - \frac{1}{2}) λ$
$2 \times 1.33 \times t = \frac{1}{2} λ$
$t = \frac{λ}{2 \times 2 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{4 \times 1.33}$
$t = \frac{532 \times 10^{- 9}}{5.32}$
$t = 100 \times 10^{- 9} m$
$t = 100 nm$

TS- EAMCET-05.05.2018

WAVE OPTICS

283183 In Young's double slit experiment, the intensity of light coming from one of the slits is double the intensity from the other slit.
The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

1 34

2 40

3 25

4 38

Explanation:

: Given that,
$I_{1} = 2 I_{2}$
We know that,
$\frac{I_{max}}{I_{min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{\frac{2 I_{2}}{I_{2}}} + 1}{\sqrt{\frac{2 I_{2}}{I_{2}}} - 1})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{1.414 + 1}{1.414 - 1})}^{2}$
$\frac{I_{max}}{I_{min}} ◻ 34 : 1$ Hence, the ratio of the maximum to the minimum intensity in the interference fringe pattern observed is
34

Manipal UGET-2018

WAVE OPTICS

283185 Two coherent sources of equal intensities produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by $50 %$ by reducing its width, then the intensity of light at the same will be units.

1 90.35

2 89.86

3 67.35

4 72.85

Explanation:

: Given,
$I_{max} = 100$
$I_{max} = I_{o} + I_{o} + 2 \sqrt{I_{0} I_{0}}$
$4 I_{0} = 100$
$I_{0} = 25$
If the intensity of one of the source is reduced by $50 %$ then, $I_{1} = 25$
$I_{2} = 25 - 25 \times \frac{50}{100}$
$I_{2} = 12.5$
$I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cdot \cos θ$
$I_{R} = 25 + 12.5 + 2 \sqrt{25 \times 12.5} \cos 0^{\circ}$
$I_{R} = 37.5 + 2 (17.67)$
$I_{R} = 37.5 + 35.35$
$I_{R} = 72.85$

AP EAMCET-25.04.2018