QBManager

WAVE OPTICS

283095 A single slit of width a is illuminated by violet light of wavelength $400 nm$ and the width of the diffraction pattern is measured as $y$ . When half of the slit width is covered and illuminated by yellow light of wavelength $600 nm$ , the width of the diffraction pattern is :

1 the pattern vanishes and the width is zero

2

y / 3

3

3 y

4 none of these

Explanation:

: Given, $λ = 400 nm, λ^{'} = 600 nm$
$a^{'} = \frac{a}{2}$
In single slit experiment width of central maximum, Width $(y) = 2 β = \frac{2 λ D}{a}$
$y \propto \frac{λ}{a}$
$\frac{y^{'}}{y} = \frac{λ^{'}}{λ} \times \frac{a}{a^{'}}$
$\frac{y^{'}}{y} = \frac{600}{400} \times \frac{a}{(a / 2)}$
$\frac{y^{'}}{y} = 3$
$y^{'} = 3 y$

Karnataka CET-2005

WAVE OPTICS

283104 The distance of moon from the earth is $3.8 \times$ $10^{5} km$ . Supposing that the eye is most sensitive to the light of wavelength $550 nm$ , the separation of two points on the moon that can be resolved by a $500 cm$ telescope is

1

50 m

2

55 m

3

51 m

4

60 m

Explanation:

: Given, $λ = 550 nm = 550 \times 10^{- 9} m$ $r = 3.8 \times 10^{5} km = 3.8 \times 10^{8} m$
$Limit of resolution (Δ θ) = \frac{1}{Resolving power (RP)}$
$But, Δ θ = \frac{Distance between points on surface of moon (x)}{Distance from telescope (r)}$
$\frac{1}{RP} = \frac{x}{r} \Rightarrow x = \frac{r}{RP} = \frac{r}{d / 1.22 λ}$
$x = \frac{1.22 r λ}{d}$
$x = \frac{1.22 \times 550 \times 10^{- 9} \times 3.8 \times 10^{8}}{500 \times 10^{- 2}} = 51 m$

J and K-CET-2013

WAVE OPTICS

283110 A beam of light of $λ = 600 nm$ from a distant source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between first dark fringes on either side of the central bright fringe is

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

: Given that, wavelength of beam light $(λ) =$ $600 nm = 600 \times 10^{- 9} m$
Diffraction $(D) = 2 m$
Slit width $(d) = 1 mm = 1 \times 10^{- 3} m$
For a dark fringe -
$\frac{d y}{D} = λ$
$y = \frac{D λ}{d} = \frac{2 \times 600 \times 10^{- 9}}{10^{- 3}} = 1.2 mm$
$∴$ Distance between the first dark fringes on either side of central bright fringe will be-
$2 y = 2 \times 1.2 = 2.4 mm$

AIPMT - 2014

WAVE OPTICS

283119 Which of the following wavelength is in the infrared region?

1

5460 \AA

2

5893 \AA

3

4320 \AA

4

7000 \AA

Explanation:

: Wavelength is in the infrared region $\to 700 \AA$
The wavelength range of infrared rays is $7 \times 10^{- 7}$ to $3 \times 10^{- 7}$
| Name | Wavelength |
| :--- | :--- |
| Gamma ray | less than $0.01 nm$ |
| X-ray | $0.01 nm - 10 nm$ |
| Ultraviolet | $10 nm - 400 nm$ |
| Visible | $400 nm - 700 nm$ |
| Infrared | $700 nm - 1 mm$ |
| Microwave | $1 mm - 1 m$ |
| Radio | $1 m - 10000 km$ |

EAMCET-1999

WAVE OPTICS

283095 A single slit of width a is illuminated by violet light of wavelength $400 nm$ and the width of the diffraction pattern is measured as $y$ . When half of the slit width is covered and illuminated by yellow light of wavelength $600 nm$ , the width of the diffraction pattern is :

1 the pattern vanishes and the width is zero

2

y / 3

3

3 y

4 none of these

Explanation:

: Given, $λ = 400 nm, λ^{'} = 600 nm$
$a^{'} = \frac{a}{2}$
In single slit experiment width of central maximum, Width $(y) = 2 β = \frac{2 λ D}{a}$
$y \propto \frac{λ}{a}$
$\frac{y^{'}}{y} = \frac{λ^{'}}{λ} \times \frac{a}{a^{'}}$
$\frac{y^{'}}{y} = \frac{600}{400} \times \frac{a}{(a / 2)}$
$\frac{y^{'}}{y} = 3$
$y^{'} = 3 y$

Karnataka CET-2005

WAVE OPTICS

283104 The distance of moon from the earth is $3.8 \times$ $10^{5} km$ . Supposing that the eye is most sensitive to the light of wavelength $550 nm$ , the separation of two points on the moon that can be resolved by a $500 cm$ telescope is

1

50 m

2

55 m

3

51 m

4

60 m

Explanation:

: Given, $λ = 550 nm = 550 \times 10^{- 9} m$ $r = 3.8 \times 10^{5} km = 3.8 \times 10^{8} m$
$Limit of resolution (Δ θ) = \frac{1}{Resolving power (RP)}$
$But, Δ θ = \frac{Distance between points on surface of moon (x)}{Distance from telescope (r)}$
$\frac{1}{RP} = \frac{x}{r} \Rightarrow x = \frac{r}{RP} = \frac{r}{d / 1.22 λ}$
$x = \frac{1.22 r λ}{d}$
$x = \frac{1.22 \times 550 \times 10^{- 9} \times 3.8 \times 10^{8}}{500 \times 10^{- 2}} = 51 m$

J and K-CET-2013

WAVE OPTICS

283110 A beam of light of $λ = 600 nm$ from a distant source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between first dark fringes on either side of the central bright fringe is

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

: Given that, wavelength of beam light $(λ) =$ $600 nm = 600 \times 10^{- 9} m$
Diffraction $(D) = 2 m$
Slit width $(d) = 1 mm = 1 \times 10^{- 3} m$
For a dark fringe -
$\frac{d y}{D} = λ$
$y = \frac{D λ}{d} = \frac{2 \times 600 \times 10^{- 9}}{10^{- 3}} = 1.2 mm$
$∴$ Distance between the first dark fringes on either side of central bright fringe will be-
$2 y = 2 \times 1.2 = 2.4 mm$

AIPMT - 2014

WAVE OPTICS

283119 Which of the following wavelength is in the infrared region?

1

5460 \AA

2

5893 \AA

3

4320 \AA

4

7000 \AA

Explanation:

: Wavelength is in the infrared region $\to 700 \AA$
The wavelength range of infrared rays is $7 \times 10^{- 7}$ to $3 \times 10^{- 7}$
| Name | Wavelength |
| :--- | :--- |
| Gamma ray | less than $0.01 nm$ |
| X-ray | $0.01 nm - 10 nm$ |
| Ultraviolet | $10 nm - 400 nm$ |
| Visible | $400 nm - 700 nm$ |
| Infrared | $700 nm - 1 mm$ |
| Microwave | $1 mm - 1 m$ |
| Radio | $1 m - 10000 km$ |

EAMCET-1999

WAVE OPTICS

283095 A single slit of width a is illuminated by violet light of wavelength $400 nm$ and the width of the diffraction pattern is measured as $y$ . When half of the slit width is covered and illuminated by yellow light of wavelength $600 nm$ , the width of the diffraction pattern is :

1 the pattern vanishes and the width is zero

2

y / 3

3

3 y

4 none of these

Explanation:

: Given, $λ = 400 nm, λ^{'} = 600 nm$
$a^{'} = \frac{a}{2}$
In single slit experiment width of central maximum, Width $(y) = 2 β = \frac{2 λ D}{a}$
$y \propto \frac{λ}{a}$
$\frac{y^{'}}{y} = \frac{λ^{'}}{λ} \times \frac{a}{a^{'}}$
$\frac{y^{'}}{y} = \frac{600}{400} \times \frac{a}{(a / 2)}$
$\frac{y^{'}}{y} = 3$
$y^{'} = 3 y$

Karnataka CET-2005

WAVE OPTICS

283104 The distance of moon from the earth is $3.8 \times$ $10^{5} km$ . Supposing that the eye is most sensitive to the light of wavelength $550 nm$ , the separation of two points on the moon that can be resolved by a $500 cm$ telescope is

1

50 m

2

55 m

3

51 m

4

60 m

Explanation:

: Given, $λ = 550 nm = 550 \times 10^{- 9} m$ $r = 3.8 \times 10^{5} km = 3.8 \times 10^{8} m$
$Limit of resolution (Δ θ) = \frac{1}{Resolving power (RP)}$
$But, Δ θ = \frac{Distance between points on surface of moon (x)}{Distance from telescope (r)}$
$\frac{1}{RP} = \frac{x}{r} \Rightarrow x = \frac{r}{RP} = \frac{r}{d / 1.22 λ}$
$x = \frac{1.22 r λ}{d}$
$x = \frac{1.22 \times 550 \times 10^{- 9} \times 3.8 \times 10^{8}}{500 \times 10^{- 2}} = 51 m$

J and K-CET-2013

WAVE OPTICS

283110 A beam of light of $λ = 600 nm$ from a distant source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between first dark fringes on either side of the central bright fringe is

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

: Given that, wavelength of beam light $(λ) =$ $600 nm = 600 \times 10^{- 9} m$
Diffraction $(D) = 2 m$
Slit width $(d) = 1 mm = 1 \times 10^{- 3} m$
For a dark fringe -
$\frac{d y}{D} = λ$
$y = \frac{D λ}{d} = \frac{2 \times 600 \times 10^{- 9}}{10^{- 3}} = 1.2 mm$
$∴$ Distance between the first dark fringes on either side of central bright fringe will be-
$2 y = 2 \times 1.2 = 2.4 mm$

AIPMT - 2014

WAVE OPTICS

283119 Which of the following wavelength is in the infrared region?

1

5460 \AA

2

5893 \AA

3

4320 \AA

4

7000 \AA

Explanation:

: Wavelength is in the infrared region $\to 700 \AA$
The wavelength range of infrared rays is $7 \times 10^{- 7}$ to $3 \times 10^{- 7}$
| Name | Wavelength |
| :--- | :--- |
| Gamma ray | less than $0.01 nm$ |
| X-ray | $0.01 nm - 10 nm$ |
| Ultraviolet | $10 nm - 400 nm$ |
| Visible | $400 nm - 700 nm$ |
| Infrared | $700 nm - 1 mm$ |
| Microwave | $1 mm - 1 m$ |
| Radio | $1 m - 10000 km$ |

EAMCET-1999

WAVE OPTICS

283095 A single slit of width a is illuminated by violet light of wavelength $400 nm$ and the width of the diffraction pattern is measured as $y$ . When half of the slit width is covered and illuminated by yellow light of wavelength $600 nm$ , the width of the diffraction pattern is :

1 the pattern vanishes and the width is zero

2

y / 3

3

3 y

4 none of these

Explanation:

: Given, $λ = 400 nm, λ^{'} = 600 nm$
$a^{'} = \frac{a}{2}$
In single slit experiment width of central maximum, Width $(y) = 2 β = \frac{2 λ D}{a}$
$y \propto \frac{λ}{a}$
$\frac{y^{'}}{y} = \frac{λ^{'}}{λ} \times \frac{a}{a^{'}}$
$\frac{y^{'}}{y} = \frac{600}{400} \times \frac{a}{(a / 2)}$
$\frac{y^{'}}{y} = 3$
$y^{'} = 3 y$

Karnataka CET-2005

WAVE OPTICS

283104 The distance of moon from the earth is $3.8 \times$ $10^{5} km$ . Supposing that the eye is most sensitive to the light of wavelength $550 nm$ , the separation of two points on the moon that can be resolved by a $500 cm$ telescope is

1

50 m

2

55 m

3

51 m

4

60 m

Explanation:

: Given, $λ = 550 nm = 550 \times 10^{- 9} m$ $r = 3.8 \times 10^{5} km = 3.8 \times 10^{8} m$
$Limit of resolution (Δ θ) = \frac{1}{Resolving power (RP)}$
$But, Δ θ = \frac{Distance between points on surface of moon (x)}{Distance from telescope (r)}$
$\frac{1}{RP} = \frac{x}{r} \Rightarrow x = \frac{r}{RP} = \frac{r}{d / 1.22 λ}$
$x = \frac{1.22 r λ}{d}$
$x = \frac{1.22 \times 550 \times 10^{- 9} \times 3.8 \times 10^{8}}{500 \times 10^{- 2}} = 51 m$

J and K-CET-2013

WAVE OPTICS

283110 A beam of light of $λ = 600 nm$ from a distant source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between first dark fringes on either side of the central bright fringe is

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

: Given that, wavelength of beam light $(λ) =$ $600 nm = 600 \times 10^{- 9} m$
Diffraction $(D) = 2 m$
Slit width $(d) = 1 mm = 1 \times 10^{- 3} m$
For a dark fringe -
$\frac{d y}{D} = λ$
$y = \frac{D λ}{d} = \frac{2 \times 600 \times 10^{- 9}}{10^{- 3}} = 1.2 mm$
$∴$ Distance between the first dark fringes on either side of central bright fringe will be-
$2 y = 2 \times 1.2 = 2.4 mm$

AIPMT - 2014

WAVE OPTICS

283119 Which of the following wavelength is in the infrared region?

1

5460 \AA

2

5893 \AA

3

4320 \AA

4

7000 \AA

Explanation:

: Wavelength is in the infrared region $\to 700 \AA$
The wavelength range of infrared rays is $7 \times 10^{- 7}$ to $3 \times 10^{- 7}$
| Name | Wavelength |
| :--- | :--- |
| Gamma ray | less than $0.01 nm$ |
| X-ray | $0.01 nm - 10 nm$ |
| Ultraviolet | $10 nm - 400 nm$ |
| Visible | $400 nm - 700 nm$ |
| Infrared | $700 nm - 1 mm$ |
| Microwave | $1 mm - 1 m$ |
| Radio | $1 m - 10000 km$ |

EAMCET-1999

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