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WAVE OPTICS

283330 In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $λ = I$ . The intensity of light at a point where the path difference becomes $λ / 3$ is

1

I / 4

2

I / 3

3

\frac{I}{2}

4 I

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) \times Δ x$
Case-I
For path difference $λ$
$ϕ = \frac{2 π}{λ} \times λ = 2 π$
So, intensity of light $(I) = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
$= 4 I_{0} \cos^{2} (2 π / 2)$
$I = 4 I_{0}$
Case-II
For path difference $λ / 3$
$ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$
Intensity of light,
$I^{'} = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) = 4 I_{0} \cos^{2} (\frac{2 π}{3 \times 2})$
$I^{'} = 4 I_{0} \times \frac{1}{4}$
$I^{'} = \frac{I}{4}$

Karnataka CET-2019

WAVE OPTICS

283331 In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths $4200 \AA$ and $5040 \AA$ . If the distance between the slits is $2.4 mm$ and the distance between the slits and the screen is $200$ $cm$ , the minimum distance from the central bright fringe to the point where the bright fringes due to both the wavelengths coincide is

1

0.7 mm

2

1.4 mm

3

2.1 mm

4

2.8 mm

Explanation:

: Let $n_{1}$ fringes of $λ_{1} = 4200 \AA$
And $n_{2}$ fringes of $λ_{2} = 5040 \AA$ wavelength are formed in a fixed distance on screen
So,
$n λ = constant$
$n_{1} λ_{1} = n_{2} λ_{2}$
For minimum distance to be coincide,
$n_{1} = n + 1, n_{2} = n$
$(n + 1) 4200 = n \times 5040$
$n = 5$
$∴ n_{1} = n + 1 = 5 + 1 = 6$
Required distance $x = \frac{n_{1} λ_{1} D}{d}$
$= \frac{6 \times 4200 \times 10^{- 10} \times 2}{2.4 \times 10^{- 3}}$
$= 2.1 \times 10^{- 3} m$
$= 2.1 mm$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283332 A Young's double slit experimental setup is immersed in water of refractive index 1.33. It has slit separation $1 mm$ and the distance between slits and screen is $1.33 m$ . If the wavelength of incident light on slits is $6300 \AA$ , then the fringe width on the screen is

1

6.3 mm

2

0.63 mm

3

0.63 m

4

6.3 m

Explanation:

: Given,
Refractive index $(μ) = 1.33, d = 1 mm = 10^{- 3} m$
$D = 1.33 m,$
Wavelength of incident light on slit $(λ) = 6300 \AA =$ $6300 \times 10^{- 10} m$
Fringe width $β = \frac{λ D}{μ d}$
$β = \frac{6300 \times 10^{- 10} \times 1.33}{1.33 \times 10^{- 3}}$
$β = 6.3 \times 10^{- 4} m$
$β = 0.63 mm$

AP EAMCET (21.04.2019) Shift-II

WAVE OPTICS

283333 In Young's double slit experiment, light of wavelength $5900 \AA$ is used. When the slits are 2 $mm$ apart, the fringe width is $1.2 mm$ . If the slit separation is increased to one and half times the previous value, then the fringe width will be

1

0.9 mm

2

0.8 mm

3

1.8 mm

4

1.6 mm

Explanation:

: Given,
Wavelength of light $(λ) = 5900 \AA = 5900 \times 10^{- 10} m$ $d = 2 mm$
Fringe width $(β) = 1.2 mm$
$d^{'} = d + \frac{d}{2} = \frac{3}{2} d = \frac{3}{2} \times 2 = 3 mm$
Fringe width $β = \frac{λ D}{d}$
$β \propto \frac{1}{d}$
$\frac{β^{'}}{β} = \frac{d}{d}$
$\frac{β^{'}}{1.2} = \frac{2}{3}$
$β^{'} = 0.8 mm$

AP EAMCET (23.04.2019) Shift-I

WAVE OPTICS

283334 In Young's double slit experiment, the two slits are separated by $0.5 cm$ and the screen is at $0.5 m$ form the slits. If 20000 bright fringes are counted per meter on the screen, then the wavelength of light used is

1

5000 \AA

2

5890 \AA

3

6000 \AA

4

5460 \AA

Explanation:

: Given,
$d = 0.5 cm = 0.5 \times 10^{- 2} m$
$D = 0.5 m$
$β = \frac{1}{20, 000}$
We know that,
$Wavelength of light (λ) = \frac{β d}{D}$
$= \frac{1 \times 0.5 \times 10^{- 2}}{20000 \times 0.5}$
$= 5000 \AA$

AP EAMCET (22.04.2019) Shift-II

WAVE OPTICS

283330 In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $λ = I$ . The intensity of light at a point where the path difference becomes $λ / 3$ is

1

I / 4

2

I / 3

3

\frac{I}{2}

4 I

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) \times Δ x$
Case-I
For path difference $λ$
$ϕ = \frac{2 π}{λ} \times λ = 2 π$
So, intensity of light $(I) = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
$= 4 I_{0} \cos^{2} (2 π / 2)$
$I = 4 I_{0}$
Case-II
For path difference $λ / 3$
$ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$
Intensity of light,
$I^{'} = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) = 4 I_{0} \cos^{2} (\frac{2 π}{3 \times 2})$
$I^{'} = 4 I_{0} \times \frac{1}{4}$
$I^{'} = \frac{I}{4}$

Karnataka CET-2019

WAVE OPTICS

283331 In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths $4200 \AA$ and $5040 \AA$ . If the distance between the slits is $2.4 mm$ and the distance between the slits and the screen is $200$ $cm$ , the minimum distance from the central bright fringe to the point where the bright fringes due to both the wavelengths coincide is

1

0.7 mm

2

1.4 mm

3

2.1 mm

4

2.8 mm

Explanation:

: Let $n_{1}$ fringes of $λ_{1} = 4200 \AA$
And $n_{2}$ fringes of $λ_{2} = 5040 \AA$ wavelength are formed in a fixed distance on screen
So,
$n λ = constant$
$n_{1} λ_{1} = n_{2} λ_{2}$
For minimum distance to be coincide,
$n_{1} = n + 1, n_{2} = n$
$(n + 1) 4200 = n \times 5040$
$n = 5$
$∴ n_{1} = n + 1 = 5 + 1 = 6$
Required distance $x = \frac{n_{1} λ_{1} D}{d}$
$= \frac{6 \times 4200 \times 10^{- 10} \times 2}{2.4 \times 10^{- 3}}$
$= 2.1 \times 10^{- 3} m$
$= 2.1 mm$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283332 A Young's double slit experimental setup is immersed in water of refractive index 1.33. It has slit separation $1 mm$ and the distance between slits and screen is $1.33 m$ . If the wavelength of incident light on slits is $6300 \AA$ , then the fringe width on the screen is

1

6.3 mm

2

0.63 mm

3

0.63 m

4

6.3 m

Explanation:

: Given,
Refractive index $(μ) = 1.33, d = 1 mm = 10^{- 3} m$
$D = 1.33 m,$
Wavelength of incident light on slit $(λ) = 6300 \AA =$ $6300 \times 10^{- 10} m$
Fringe width $β = \frac{λ D}{μ d}$
$β = \frac{6300 \times 10^{- 10} \times 1.33}{1.33 \times 10^{- 3}}$
$β = 6.3 \times 10^{- 4} m$
$β = 0.63 mm$

AP EAMCET (21.04.2019) Shift-II

WAVE OPTICS

283333 In Young's double slit experiment, light of wavelength $5900 \AA$ is used. When the slits are 2 $mm$ apart, the fringe width is $1.2 mm$ . If the slit separation is increased to one and half times the previous value, then the fringe width will be

1

0.9 mm

2

0.8 mm

3

1.8 mm

4

1.6 mm

Explanation:

: Given,
Wavelength of light $(λ) = 5900 \AA = 5900 \times 10^{- 10} m$ $d = 2 mm$
Fringe width $(β) = 1.2 mm$
$d^{'} = d + \frac{d}{2} = \frac{3}{2} d = \frac{3}{2} \times 2 = 3 mm$
Fringe width $β = \frac{λ D}{d}$
$β \propto \frac{1}{d}$
$\frac{β^{'}}{β} = \frac{d}{d}$
$\frac{β^{'}}{1.2} = \frac{2}{3}$
$β^{'} = 0.8 mm$

AP EAMCET (23.04.2019) Shift-I

WAVE OPTICS

283334 In Young's double slit experiment, the two slits are separated by $0.5 cm$ and the screen is at $0.5 m$ form the slits. If 20000 bright fringes are counted per meter on the screen, then the wavelength of light used is

1

5000 \AA

2

5890 \AA

3

6000 \AA

4

5460 \AA

Explanation:

: Given,
$d = 0.5 cm = 0.5 \times 10^{- 2} m$
$D = 0.5 m$
$β = \frac{1}{20, 000}$
We know that,
$Wavelength of light (λ) = \frac{β d}{D}$
$= \frac{1 \times 0.5 \times 10^{- 2}}{20000 \times 0.5}$
$= 5000 \AA$

AP EAMCET (22.04.2019) Shift-II

WAVE OPTICS

283330 In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $λ = I$ . The intensity of light at a point where the path difference becomes $λ / 3$ is

1

I / 4

2

I / 3

3

\frac{I}{2}

4 I

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) \times Δ x$
Case-I
For path difference $λ$
$ϕ = \frac{2 π}{λ} \times λ = 2 π$
So, intensity of light $(I) = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
$= 4 I_{0} \cos^{2} (2 π / 2)$
$I = 4 I_{0}$
Case-II
For path difference $λ / 3$
$ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$
Intensity of light,
$I^{'} = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) = 4 I_{0} \cos^{2} (\frac{2 π}{3 \times 2})$
$I^{'} = 4 I_{0} \times \frac{1}{4}$
$I^{'} = \frac{I}{4}$

Karnataka CET-2019

WAVE OPTICS

283331 In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths $4200 \AA$ and $5040 \AA$ . If the distance between the slits is $2.4 mm$ and the distance between the slits and the screen is $200$ $cm$ , the minimum distance from the central bright fringe to the point where the bright fringes due to both the wavelengths coincide is

1

0.7 mm

2

1.4 mm

3

2.1 mm

4

2.8 mm

Explanation:

: Let $n_{1}$ fringes of $λ_{1} = 4200 \AA$
And $n_{2}$ fringes of $λ_{2} = 5040 \AA$ wavelength are formed in a fixed distance on screen
So,
$n λ = constant$
$n_{1} λ_{1} = n_{2} λ_{2}$
For minimum distance to be coincide,
$n_{1} = n + 1, n_{2} = n$
$(n + 1) 4200 = n \times 5040$
$n = 5$
$∴ n_{1} = n + 1 = 5 + 1 = 6$
Required distance $x = \frac{n_{1} λ_{1} D}{d}$
$= \frac{6 \times 4200 \times 10^{- 10} \times 2}{2.4 \times 10^{- 3}}$
$= 2.1 \times 10^{- 3} m$
$= 2.1 mm$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283332 A Young's double slit experimental setup is immersed in water of refractive index 1.33. It has slit separation $1 mm$ and the distance between slits and screen is $1.33 m$ . If the wavelength of incident light on slits is $6300 \AA$ , then the fringe width on the screen is

1

6.3 mm

2

0.63 mm

3

0.63 m

4

6.3 m

Explanation:

: Given,
Refractive index $(μ) = 1.33, d = 1 mm = 10^{- 3} m$
$D = 1.33 m,$
Wavelength of incident light on slit $(λ) = 6300 \AA =$ $6300 \times 10^{- 10} m$
Fringe width $β = \frac{λ D}{μ d}$
$β = \frac{6300 \times 10^{- 10} \times 1.33}{1.33 \times 10^{- 3}}$
$β = 6.3 \times 10^{- 4} m$
$β = 0.63 mm$

AP EAMCET (21.04.2019) Shift-II

WAVE OPTICS

283333 In Young's double slit experiment, light of wavelength $5900 \AA$ is used. When the slits are 2 $mm$ apart, the fringe width is $1.2 mm$ . If the slit separation is increased to one and half times the previous value, then the fringe width will be

1

0.9 mm

2

0.8 mm

3

1.8 mm

4

1.6 mm

Explanation:

: Given,
Wavelength of light $(λ) = 5900 \AA = 5900 \times 10^{- 10} m$ $d = 2 mm$
Fringe width $(β) = 1.2 mm$
$d^{'} = d + \frac{d}{2} = \frac{3}{2} d = \frac{3}{2} \times 2 = 3 mm$
Fringe width $β = \frac{λ D}{d}$
$β \propto \frac{1}{d}$
$\frac{β^{'}}{β} = \frac{d}{d}$
$\frac{β^{'}}{1.2} = \frac{2}{3}$
$β^{'} = 0.8 mm$

AP EAMCET (23.04.2019) Shift-I

WAVE OPTICS

283334 In Young's double slit experiment, the two slits are separated by $0.5 cm$ and the screen is at $0.5 m$ form the slits. If 20000 bright fringes are counted per meter on the screen, then the wavelength of light used is

1

5000 \AA

2

5890 \AA

3

6000 \AA

4

5460 \AA

Explanation:

: Given,
$d = 0.5 cm = 0.5 \times 10^{- 2} m$
$D = 0.5 m$
$β = \frac{1}{20, 000}$
We know that,
$Wavelength of light (λ) = \frac{β d}{D}$
$= \frac{1 \times 0.5 \times 10^{- 2}}{20000 \times 0.5}$
$= 5000 \AA$

AP EAMCET (22.04.2019) Shift-II

WAVE OPTICS

283330 In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $λ = I$ . The intensity of light at a point where the path difference becomes $λ / 3$ is

1

I / 4

2

I / 3

3

\frac{I}{2}

4 I

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) \times Δ x$
Case-I
For path difference $λ$
$ϕ = \frac{2 π}{λ} \times λ = 2 π$
So, intensity of light $(I) = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
$= 4 I_{0} \cos^{2} (2 π / 2)$
$I = 4 I_{0}$
Case-II
For path difference $λ / 3$
$ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$
Intensity of light,
$I^{'} = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) = 4 I_{0} \cos^{2} (\frac{2 π}{3 \times 2})$
$I^{'} = 4 I_{0} \times \frac{1}{4}$
$I^{'} = \frac{I}{4}$

Karnataka CET-2019

WAVE OPTICS

283331 In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths $4200 \AA$ and $5040 \AA$ . If the distance between the slits is $2.4 mm$ and the distance between the slits and the screen is $200$ $cm$ , the minimum distance from the central bright fringe to the point where the bright fringes due to both the wavelengths coincide is

1

0.7 mm

2

1.4 mm

3

2.1 mm

4

2.8 mm

Explanation:

: Let $n_{1}$ fringes of $λ_{1} = 4200 \AA$
And $n_{2}$ fringes of $λ_{2} = 5040 \AA$ wavelength are formed in a fixed distance on screen
So,
$n λ = constant$
$n_{1} λ_{1} = n_{2} λ_{2}$
For minimum distance to be coincide,
$n_{1} = n + 1, n_{2} = n$
$(n + 1) 4200 = n \times 5040$
$n = 5$
$∴ n_{1} = n + 1 = 5 + 1 = 6$
Required distance $x = \frac{n_{1} λ_{1} D}{d}$
$= \frac{6 \times 4200 \times 10^{- 10} \times 2}{2.4 \times 10^{- 3}}$
$= 2.1 \times 10^{- 3} m$
$= 2.1 mm$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283332 A Young's double slit experimental setup is immersed in water of refractive index 1.33. It has slit separation $1 mm$ and the distance between slits and screen is $1.33 m$ . If the wavelength of incident light on slits is $6300 \AA$ , then the fringe width on the screen is

1

6.3 mm

2

0.63 mm

3

0.63 m

4

6.3 m

Explanation:

: Given,
Refractive index $(μ) = 1.33, d = 1 mm = 10^{- 3} m$
$D = 1.33 m,$
Wavelength of incident light on slit $(λ) = 6300 \AA =$ $6300 \times 10^{- 10} m$
Fringe width $β = \frac{λ D}{μ d}$
$β = \frac{6300 \times 10^{- 10} \times 1.33}{1.33 \times 10^{- 3}}$
$β = 6.3 \times 10^{- 4} m$
$β = 0.63 mm$

AP EAMCET (21.04.2019) Shift-II

WAVE OPTICS

283333 In Young's double slit experiment, light of wavelength $5900 \AA$ is used. When the slits are 2 $mm$ apart, the fringe width is $1.2 mm$ . If the slit separation is increased to one and half times the previous value, then the fringe width will be

1

0.9 mm

2

0.8 mm

3

1.8 mm

4

1.6 mm

Explanation:

: Given,
Wavelength of light $(λ) = 5900 \AA = 5900 \times 10^{- 10} m$ $d = 2 mm$
Fringe width $(β) = 1.2 mm$
$d^{'} = d + \frac{d}{2} = \frac{3}{2} d = \frac{3}{2} \times 2 = 3 mm$
Fringe width $β = \frac{λ D}{d}$
$β \propto \frac{1}{d}$
$\frac{β^{'}}{β} = \frac{d}{d}$
$\frac{β^{'}}{1.2} = \frac{2}{3}$
$β^{'} = 0.8 mm$

AP EAMCET (23.04.2019) Shift-I

WAVE OPTICS

283334 In Young's double slit experiment, the two slits are separated by $0.5 cm$ and the screen is at $0.5 m$ form the slits. If 20000 bright fringes are counted per meter on the screen, then the wavelength of light used is

1

5000 \AA

2

5890 \AA

3

6000 \AA

4

5460 \AA

Explanation:

: Given,
$d = 0.5 cm = 0.5 \times 10^{- 2} m$
$D = 0.5 m$
$β = \frac{1}{20, 000}$
We know that,
$Wavelength of light (λ) = \frac{β d}{D}$
$= \frac{1 \times 0.5 \times 10^{- 2}}{20000 \times 0.5}$
$= 5000 \AA$

AP EAMCET (22.04.2019) Shift-II

WAVE OPTICS

283330 In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $λ = I$ . The intensity of light at a point where the path difference becomes $λ / 3$ is

1

I / 4

2

I / 3

3

\frac{I}{2}

4 I

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) \times Δ x$
Case-I
For path difference $λ$
$ϕ = \frac{2 π}{λ} \times λ = 2 π$
So, intensity of light $(I) = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
$= 4 I_{0} \cos^{2} (2 π / 2)$
$I = 4 I_{0}$
Case-II
For path difference $λ / 3$
$ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$
Intensity of light,
$I^{'} = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) = 4 I_{0} \cos^{2} (\frac{2 π}{3 \times 2})$
$I^{'} = 4 I_{0} \times \frac{1}{4}$
$I^{'} = \frac{I}{4}$

Karnataka CET-2019

WAVE OPTICS

283331 In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths $4200 \AA$ and $5040 \AA$ . If the distance between the slits is $2.4 mm$ and the distance between the slits and the screen is $200$ $cm$ , the minimum distance from the central bright fringe to the point where the bright fringes due to both the wavelengths coincide is

1

0.7 mm

2

1.4 mm

3

2.1 mm

4

2.8 mm

Explanation:

: Let $n_{1}$ fringes of $λ_{1} = 4200 \AA$
And $n_{2}$ fringes of $λ_{2} = 5040 \AA$ wavelength are formed in a fixed distance on screen
So,
$n λ = constant$
$n_{1} λ_{1} = n_{2} λ_{2}$
For minimum distance to be coincide,
$n_{1} = n + 1, n_{2} = n$
$(n + 1) 4200 = n \times 5040$
$n = 5$
$∴ n_{1} = n + 1 = 5 + 1 = 6$
Required distance $x = \frac{n_{1} λ_{1} D}{d}$
$= \frac{6 \times 4200 \times 10^{- 10} \times 2}{2.4 \times 10^{- 3}}$
$= 2.1 \times 10^{- 3} m$
$= 2.1 mm$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283332 A Young's double slit experimental setup is immersed in water of refractive index 1.33. It has slit separation $1 mm$ and the distance between slits and screen is $1.33 m$ . If the wavelength of incident light on slits is $6300 \AA$ , then the fringe width on the screen is

1

6.3 mm

2

0.63 mm

3

0.63 m

4

6.3 m

Explanation:

: Given,
Refractive index $(μ) = 1.33, d = 1 mm = 10^{- 3} m$
$D = 1.33 m,$
Wavelength of incident light on slit $(λ) = 6300 \AA =$ $6300 \times 10^{- 10} m$
Fringe width $β = \frac{λ D}{μ d}$
$β = \frac{6300 \times 10^{- 10} \times 1.33}{1.33 \times 10^{- 3}}$
$β = 6.3 \times 10^{- 4} m$
$β = 0.63 mm$

AP EAMCET (21.04.2019) Shift-II

WAVE OPTICS

283333 In Young's double slit experiment, light of wavelength $5900 \AA$ is used. When the slits are 2 $mm$ apart, the fringe width is $1.2 mm$ . If the slit separation is increased to one and half times the previous value, then the fringe width will be

1

0.9 mm

2

0.8 mm

3

1.8 mm

4

1.6 mm

Explanation:

: Given,
Wavelength of light $(λ) = 5900 \AA = 5900 \times 10^{- 10} m$ $d = 2 mm$
Fringe width $(β) = 1.2 mm$
$d^{'} = d + \frac{d}{2} = \frac{3}{2} d = \frac{3}{2} \times 2 = 3 mm$
Fringe width $β = \frac{λ D}{d}$
$β \propto \frac{1}{d}$
$\frac{β^{'}}{β} = \frac{d}{d}$
$\frac{β^{'}}{1.2} = \frac{2}{3}$
$β^{'} = 0.8 mm$

AP EAMCET (23.04.2019) Shift-I

WAVE OPTICS

283334 In Young's double slit experiment, the two slits are separated by $0.5 cm$ and the screen is at $0.5 m$ form the slits. If 20000 bright fringes are counted per meter on the screen, then the wavelength of light used is

1

5000 \AA

2

5890 \AA

3

6000 \AA

4

5460 \AA

Explanation:

: Given,
$d = 0.5 cm = 0.5 \times 10^{- 2} m$
$D = 0.5 m$
$β = \frac{1}{20, 000}$
We know that,
$Wavelength of light (λ) = \frac{β d}{D}$
$= \frac{1 \times 0.5 \times 10^{- 2}}{20000 \times 0.5}$
$= 5000 \AA$

AP EAMCET (22.04.2019) Shift-II

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