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PHXII10:WAVE OPTICS

367907 Diameter of human eye lens is $2 m m$ . What will be the minimum distance between two points to resolve them, which are situated at a distance of $50 m$ from eye? The wavelength of light is $5000 \overset{\circ}{A} .$

1

2.32 m

2

4.28 m m

3

1.25 c m

4

12.48 c m

Explanation:

Angular limit of resolution of eye
$= \frac{Wave length of light}{Diameter of eye lens}$
$i . e . θ = \frac{λ}{d} (1)$
If $y$ is the minimum resolution between
two objects at distance $D$ from eye, then
$θ = \frac{y}{D} (2)$
From Eqs. (1) and (2), we have
$\frac{y}{D} = \frac{λ}{d} or y = \frac{λ D}{d} (3)$
Given, $λ = 5000 \overset{\circ}{A} = 5 \times 10^{- 7} m,$
$D = 50 m$
and $d = 2 m m = 2 \times 10^{- 3} m$
substituting in Eq. (3), we get
$y = \frac{5 \times 10^{- 7} \times 50}{2 \times 10^{- 3}} = 12.5 \times 10^{- 3} m$
$= 1.25 c m$

PHXII10:WAVE OPTICS

367908 The head lights of a jeep are 1.2 $m$ apart. If the pupil of the eye of an observer has a diameter of 2 $m m$ and light of wavelength $5896 \overset{\circ}{A}$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated ?

1

33.9 k m

2

33.9 m

3

3.34 k m

4

3.36 m

Explanation:

Distance of jeep $x = \frac{D \times d}{1.22 \times λ}$
$D$ = Diameter of lens, $d$ = separation between
sources.
$∴ x = \frac{(2 \times 10^{- 3}) \times 1.2}{1.22 \times 5896 \times 10^{- 10}} \approx 3337 m \approx 3.34 k m$

PHXII10:WAVE OPTICS

367909 Two parallel pillars are $11 k m$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

1

20.8 m

2

3.2 m

3

183 m

4

91.5 m

Explanation:

Limit of resolution of eye is ${(\frac{1}{60})}^{\circ}$ the pillars will be seen distinctly if $d θ > {(\frac{1}{60})}^{\circ}$
i.e., $\frac{d}{x} > (\frac{1}{60}) \times \frac{π}{180}$
Where $x$ is the distance between the pillars and the eye and $d$ is the distance between two pillars.
$\Rightarrow d > \frac{π \times x}{60 \times 180}$
$\Rightarrow d > \frac{3.14 \times 11 \times 10^{3}}{60 \times 180} \Rightarrow d > 3.2 m$

PHXII10:WAVE OPTICS

367910 Two luminous point sources separated by a certain distance are at 10 $k m$ form an observe. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is 500 $n m$ , the distance of separation between the point sources just seen to be resolved is

1

12.2 m

2

24.2 m

3

2.44 m

4

1.22 m

Explanation:

According to Rayeigh’s criterion, minimum angular separation $θ = \frac{1.22 λ}{d_{e}}$
where $λ$ = wavelength of light, $d_{e}$ = diameter of the pupil of the eye.
$θ = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 2.44 \times 10^{- 4} r a d i a n .$
Now $w$ , $θ = \frac{a}{D}$
Where, $a$ = distance of separation between two points and
$D$ = distance of observer from point sources
$∴ a = D \times θ = 10 \times 10^{3} \times 2.44 \times 10^{- 4} = 2.44 m$
supporting img

KCET - 2009

PHXII10:WAVE OPTICS

367911 An astronaut is looking down on earth’s surface from a space shuttle at an altitude of 400 $k m$ . Assuming that the astronaut’s pupil diameter is 5 $m m$ and the wavelength of visible light is 500 $n m$ . The astronaut will be able to resolve linear object of the size

1

5 m

2

0.5 m

3

500 m

4

50 m

Explanation:

$\frac{x}{d} = \frac{1.22 λ}{a} \Rightarrow x = \frac{1.22 λ d}{a}$
$= \frac{1.22 \times 500 \times 10^{- 9} \times 400 \times 10^{3}}{5 \times 10^{- 3}} = 50 m$

PHXII10:WAVE OPTICS

367907 Diameter of human eye lens is $2 m m$ . What will be the minimum distance between two points to resolve them, which are situated at a distance of $50 m$ from eye? The wavelength of light is $5000 \overset{\circ}{A} .$

1

2.32 m

2

4.28 m m

3

1.25 c m

4

12.48 c m

Explanation:

Angular limit of resolution of eye
$= \frac{Wave length of light}{Diameter of eye lens}$
$i . e . θ = \frac{λ}{d} (1)$
If $y$ is the minimum resolution between
two objects at distance $D$ from eye, then
$θ = \frac{y}{D} (2)$
From Eqs. (1) and (2), we have
$\frac{y}{D} = \frac{λ}{d} or y = \frac{λ D}{d} (3)$
Given, $λ = 5000 \overset{\circ}{A} = 5 \times 10^{- 7} m,$
$D = 50 m$
and $d = 2 m m = 2 \times 10^{- 3} m$
substituting in Eq. (3), we get
$y = \frac{5 \times 10^{- 7} \times 50}{2 \times 10^{- 3}} = 12.5 \times 10^{- 3} m$
$= 1.25 c m$

PHXII10:WAVE OPTICS

367908 The head lights of a jeep are 1.2 $m$ apart. If the pupil of the eye of an observer has a diameter of 2 $m m$ and light of wavelength $5896 \overset{\circ}{A}$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated ?

1

33.9 k m

2

33.9 m

3

3.34 k m

4

3.36 m

Explanation:

Distance of jeep $x = \frac{D \times d}{1.22 \times λ}$
$D$ = Diameter of lens, $d$ = separation between
sources.
$∴ x = \frac{(2 \times 10^{- 3}) \times 1.2}{1.22 \times 5896 \times 10^{- 10}} \approx 3337 m \approx 3.34 k m$

PHXII10:WAVE OPTICS

367909 Two parallel pillars are $11 k m$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

1

20.8 m

2

3.2 m

3

183 m

4

91.5 m

Explanation:

Limit of resolution of eye is ${(\frac{1}{60})}^{\circ}$ the pillars will be seen distinctly if $d θ > {(\frac{1}{60})}^{\circ}$
i.e., $\frac{d}{x} > (\frac{1}{60}) \times \frac{π}{180}$
Where $x$ is the distance between the pillars and the eye and $d$ is the distance between two pillars.
$\Rightarrow d > \frac{π \times x}{60 \times 180}$
$\Rightarrow d > \frac{3.14 \times 11 \times 10^{3}}{60 \times 180} \Rightarrow d > 3.2 m$

PHXII10:WAVE OPTICS

367910 Two luminous point sources separated by a certain distance are at 10 $k m$ form an observe. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is 500 $n m$ , the distance of separation between the point sources just seen to be resolved is

1

12.2 m

2

24.2 m

3

2.44 m

4

1.22 m

Explanation:

According to Rayeigh’s criterion, minimum angular separation $θ = \frac{1.22 λ}{d_{e}}$
where $λ$ = wavelength of light, $d_{e}$ = diameter of the pupil of the eye.
$θ = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 2.44 \times 10^{- 4} r a d i a n .$
Now $w$ , $θ = \frac{a}{D}$
Where, $a$ = distance of separation between two points and
$D$ = distance of observer from point sources
$∴ a = D \times θ = 10 \times 10^{3} \times 2.44 \times 10^{- 4} = 2.44 m$
supporting img

KCET - 2009

PHXII10:WAVE OPTICS

367911 An astronaut is looking down on earth’s surface from a space shuttle at an altitude of 400 $k m$ . Assuming that the astronaut’s pupil diameter is 5 $m m$ and the wavelength of visible light is 500 $n m$ . The astronaut will be able to resolve linear object of the size

1

5 m

2

0.5 m

3

500 m

4

50 m

Explanation:

$\frac{x}{d} = \frac{1.22 λ}{a} \Rightarrow x = \frac{1.22 λ d}{a}$
$= \frac{1.22 \times 500 \times 10^{- 9} \times 400 \times 10^{3}}{5 \times 10^{- 3}} = 50 m$

PHXII10:WAVE OPTICS

367907 Diameter of human eye lens is $2 m m$ . What will be the minimum distance between two points to resolve them, which are situated at a distance of $50 m$ from eye? The wavelength of light is $5000 \overset{\circ}{A} .$

1

2.32 m

2

4.28 m m

3

1.25 c m

4

12.48 c m

Explanation:

Angular limit of resolution of eye
$= \frac{Wave length of light}{Diameter of eye lens}$
$i . e . θ = \frac{λ}{d} (1)$
If $y$ is the minimum resolution between
two objects at distance $D$ from eye, then
$θ = \frac{y}{D} (2)$
From Eqs. (1) and (2), we have
$\frac{y}{D} = \frac{λ}{d} or y = \frac{λ D}{d} (3)$
Given, $λ = 5000 \overset{\circ}{A} = 5 \times 10^{- 7} m,$
$D = 50 m$
and $d = 2 m m = 2 \times 10^{- 3} m$
substituting in Eq. (3), we get
$y = \frac{5 \times 10^{- 7} \times 50}{2 \times 10^{- 3}} = 12.5 \times 10^{- 3} m$
$= 1.25 c m$

PHXII10:WAVE OPTICS

367908 The head lights of a jeep are 1.2 $m$ apart. If the pupil of the eye of an observer has a diameter of 2 $m m$ and light of wavelength $5896 \overset{\circ}{A}$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated ?

1

33.9 k m

2

33.9 m

3

3.34 k m

4

3.36 m

Explanation:

Distance of jeep $x = \frac{D \times d}{1.22 \times λ}$
$D$ = Diameter of lens, $d$ = separation between
sources.
$∴ x = \frac{(2 \times 10^{- 3}) \times 1.2}{1.22 \times 5896 \times 10^{- 10}} \approx 3337 m \approx 3.34 k m$

PHXII10:WAVE OPTICS

367909 Two parallel pillars are $11 k m$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

1

20.8 m

2

3.2 m

3

183 m

4

91.5 m

Explanation:

Limit of resolution of eye is ${(\frac{1}{60})}^{\circ}$ the pillars will be seen distinctly if $d θ > {(\frac{1}{60})}^{\circ}$
i.e., $\frac{d}{x} > (\frac{1}{60}) \times \frac{π}{180}$
Where $x$ is the distance between the pillars and the eye and $d$ is the distance between two pillars.
$\Rightarrow d > \frac{π \times x}{60 \times 180}$
$\Rightarrow d > \frac{3.14 \times 11 \times 10^{3}}{60 \times 180} \Rightarrow d > 3.2 m$

PHXII10:WAVE OPTICS

367910 Two luminous point sources separated by a certain distance are at 10 $k m$ form an observe. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is 500 $n m$ , the distance of separation between the point sources just seen to be resolved is

1

12.2 m

2

24.2 m

3

2.44 m

4

1.22 m

Explanation:

According to Rayeigh’s criterion, minimum angular separation $θ = \frac{1.22 λ}{d_{e}}$
where $λ$ = wavelength of light, $d_{e}$ = diameter of the pupil of the eye.
$θ = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 2.44 \times 10^{- 4} r a d i a n .$
Now $w$ , $θ = \frac{a}{D}$
Where, $a$ = distance of separation between two points and
$D$ = distance of observer from point sources
$∴ a = D \times θ = 10 \times 10^{3} \times 2.44 \times 10^{- 4} = 2.44 m$
supporting img

KCET - 2009

PHXII10:WAVE OPTICS

367911 An astronaut is looking down on earth’s surface from a space shuttle at an altitude of 400 $k m$ . Assuming that the astronaut’s pupil diameter is 5 $m m$ and the wavelength of visible light is 500 $n m$ . The astronaut will be able to resolve linear object of the size

1

5 m

2

0.5 m

3

500 m

4

50 m

Explanation:

$\frac{x}{d} = \frac{1.22 λ}{a} \Rightarrow x = \frac{1.22 λ d}{a}$
$= \frac{1.22 \times 500 \times 10^{- 9} \times 400 \times 10^{3}}{5 \times 10^{- 3}} = 50 m$

PHXII10:WAVE OPTICS

367907 Diameter of human eye lens is $2 m m$ . What will be the minimum distance between two points to resolve them, which are situated at a distance of $50 m$ from eye? The wavelength of light is $5000 \overset{\circ}{A} .$

1

2.32 m

2

4.28 m m

3

1.25 c m

4

12.48 c m

Explanation:

Angular limit of resolution of eye
$= \frac{Wave length of light}{Diameter of eye lens}$
$i . e . θ = \frac{λ}{d} (1)$
If $y$ is the minimum resolution between
two objects at distance $D$ from eye, then
$θ = \frac{y}{D} (2)$
From Eqs. (1) and (2), we have
$\frac{y}{D} = \frac{λ}{d} or y = \frac{λ D}{d} (3)$
Given, $λ = 5000 \overset{\circ}{A} = 5 \times 10^{- 7} m,$
$D = 50 m$
and $d = 2 m m = 2 \times 10^{- 3} m$
substituting in Eq. (3), we get
$y = \frac{5 \times 10^{- 7} \times 50}{2 \times 10^{- 3}} = 12.5 \times 10^{- 3} m$
$= 1.25 c m$

PHXII10:WAVE OPTICS

367908 The head lights of a jeep are 1.2 $m$ apart. If the pupil of the eye of an observer has a diameter of 2 $m m$ and light of wavelength $5896 \overset{\circ}{A}$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated ?

1

33.9 k m

2

33.9 m

3

3.34 k m

4

3.36 m

Explanation:

Distance of jeep $x = \frac{D \times d}{1.22 \times λ}$
$D$ = Diameter of lens, $d$ = separation between
sources.
$∴ x = \frac{(2 \times 10^{- 3}) \times 1.2}{1.22 \times 5896 \times 10^{- 10}} \approx 3337 m \approx 3.34 k m$

PHXII10:WAVE OPTICS

367909 Two parallel pillars are $11 k m$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

1

20.8 m

2

3.2 m

3

183 m

4

91.5 m

Explanation:

Limit of resolution of eye is ${(\frac{1}{60})}^{\circ}$ the pillars will be seen distinctly if $d θ > {(\frac{1}{60})}^{\circ}$
i.e., $\frac{d}{x} > (\frac{1}{60}) \times \frac{π}{180}$
Where $x$ is the distance between the pillars and the eye and $d$ is the distance between two pillars.
$\Rightarrow d > \frac{π \times x}{60 \times 180}$
$\Rightarrow d > \frac{3.14 \times 11 \times 10^{3}}{60 \times 180} \Rightarrow d > 3.2 m$

PHXII10:WAVE OPTICS

367910 Two luminous point sources separated by a certain distance are at 10 $k m$ form an observe. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is 500 $n m$ , the distance of separation between the point sources just seen to be resolved is

1

12.2 m

2

24.2 m

3

2.44 m

4

1.22 m

Explanation:

According to Rayeigh’s criterion, minimum angular separation $θ = \frac{1.22 λ}{d_{e}}$
where $λ$ = wavelength of light, $d_{e}$ = diameter of the pupil of the eye.
$θ = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 2.44 \times 10^{- 4} r a d i a n .$
Now $w$ , $θ = \frac{a}{D}$
Where, $a$ = distance of separation between two points and
$D$ = distance of observer from point sources
$∴ a = D \times θ = 10 \times 10^{3} \times 2.44 \times 10^{- 4} = 2.44 m$
supporting img

KCET - 2009

PHXII10:WAVE OPTICS

367911 An astronaut is looking down on earth’s surface from a space shuttle at an altitude of 400 $k m$ . Assuming that the astronaut’s pupil diameter is 5 $m m$ and the wavelength of visible light is 500 $n m$ . The astronaut will be able to resolve linear object of the size

1

5 m

2

0.5 m

3

500 m

4

50 m

Explanation:

$\frac{x}{d} = \frac{1.22 λ}{a} \Rightarrow x = \frac{1.22 λ d}{a}$
$= \frac{1.22 \times 500 \times 10^{- 9} \times 400 \times 10^{3}}{5 \times 10^{- 3}} = 50 m$

PHXII10:WAVE OPTICS

367907 Diameter of human eye lens is $2 m m$ . What will be the minimum distance between two points to resolve them, which are situated at a distance of $50 m$ from eye? The wavelength of light is $5000 \overset{\circ}{A} .$

1

2.32 m

2

4.28 m m

3

1.25 c m

4

12.48 c m

Explanation:

Angular limit of resolution of eye
$= \frac{Wave length of light}{Diameter of eye lens}$
$i . e . θ = \frac{λ}{d} (1)$
If $y$ is the minimum resolution between
two objects at distance $D$ from eye, then
$θ = \frac{y}{D} (2)$
From Eqs. (1) and (2), we have
$\frac{y}{D} = \frac{λ}{d} or y = \frac{λ D}{d} (3)$
Given, $λ = 5000 \overset{\circ}{A} = 5 \times 10^{- 7} m,$
$D = 50 m$
and $d = 2 m m = 2 \times 10^{- 3} m$
substituting in Eq. (3), we get
$y = \frac{5 \times 10^{- 7} \times 50}{2 \times 10^{- 3}} = 12.5 \times 10^{- 3} m$
$= 1.25 c m$

PHXII10:WAVE OPTICS

367908 The head lights of a jeep are 1.2 $m$ apart. If the pupil of the eye of an observer has a diameter of 2 $m m$ and light of wavelength $5896 \overset{\circ}{A}$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated ?

1

33.9 k m

2

33.9 m

3

3.34 k m

4

3.36 m

Explanation:

Distance of jeep $x = \frac{D \times d}{1.22 \times λ}$
$D$ = Diameter of lens, $d$ = separation between
sources.
$∴ x = \frac{(2 \times 10^{- 3}) \times 1.2}{1.22 \times 5896 \times 10^{- 10}} \approx 3337 m \approx 3.34 k m$

PHXII10:WAVE OPTICS

367909 Two parallel pillars are $11 k m$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

1

20.8 m

2

3.2 m

3

183 m

4

91.5 m

Explanation:

Limit of resolution of eye is ${(\frac{1}{60})}^{\circ}$ the pillars will be seen distinctly if $d θ > {(\frac{1}{60})}^{\circ}$
i.e., $\frac{d}{x} > (\frac{1}{60}) \times \frac{π}{180}$
Where $x$ is the distance between the pillars and the eye and $d$ is the distance between two pillars.
$\Rightarrow d > \frac{π \times x}{60 \times 180}$
$\Rightarrow d > \frac{3.14 \times 11 \times 10^{3}}{60 \times 180} \Rightarrow d > 3.2 m$

PHXII10:WAVE OPTICS

367910 Two luminous point sources separated by a certain distance are at 10 $k m$ form an observe. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is 500 $n m$ , the distance of separation between the point sources just seen to be resolved is

1

12.2 m

2

24.2 m

3

2.44 m

4

1.22 m

Explanation:

According to Rayeigh’s criterion, minimum angular separation $θ = \frac{1.22 λ}{d_{e}}$
where $λ$ = wavelength of light, $d_{e}$ = diameter of the pupil of the eye.
$θ = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 2.44 \times 10^{- 4} r a d i a n .$
Now $w$ , $θ = \frac{a}{D}$
Where, $a$ = distance of separation between two points and
$D$ = distance of observer from point sources
$∴ a = D \times θ = 10 \times 10^{3} \times 2.44 \times 10^{- 4} = 2.44 m$
supporting img

KCET - 2009

PHXII10:WAVE OPTICS

367911 An astronaut is looking down on earth’s surface from a space shuttle at an altitude of 400 $k m$ . Assuming that the astronaut’s pupil diameter is 5 $m m$ and the wavelength of visible light is 500 $n m$ . The astronaut will be able to resolve linear object of the size

1

5 m

2

0.5 m

3

500 m

4

50 m

Explanation:

$\frac{x}{d} = \frac{1.22 λ}{a} \Rightarrow x = \frac{1.22 λ d}{a}$
$= \frac{1.22 \times 500 \times 10^{- 9} \times 400 \times 10^{3}}{5 \times 10^{- 3}} = 50 m$