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Ray Optics

282915 Two luminous point sources separated by a certain distance are at $10 km$ from an observer. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is $500 nm$ , 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:

C: Given,
$\begin{array}{r} λ_{1} = 500 nm = 500 \times 10^{- 9} m \\ d = 2.5 \times 10^{- 3} m \\ r = 10 km = 10, 000 m \end{array}$
According to Rayleigh's criterion,
Using
$\begin{array}{r} α = \frac{1.22 λ}{d} \\ α = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} \\ α = 2.44 \times 10^{- 4} \\ α = \frac{l}{r} \end{array}$
But,
$α = \frac{l}{r}$
Distance of separation $(l) = α . r$
$\begin{array}{r} l = 2.44 \times 10^{- 4} \times 10, 000 \\ l = 2.44 m \end{array}$

Manipal UGET-2009

Ray Optics

282916 In a telescope, the focal length of objective is 60 $cm$ and eye piece is $10 cm$ . When it is focused on an object parallel rays come out of eyepiece. If the angle subtended at objective by object is $3^{\circ}$ , then what is the angular width of image?

1

24^{\circ}

2

18^{\circ}

3

12^{\circ}

4

6^{\circ}

Explanation:

B: Given,
$f_{o} = 60 cm, f_{e} = 10 cm$
$α = 3^{\circ}$
We know that -
Magnification, $\frac{β}{α} = \frac{f_{o}}{f_{e}}$
Where, $β =$ angular width of image
$α =$ angle subtended at object
$f_{o} =$ focal length of objective
$f_{e} =$ focal length of eye piece
$β = 3^{\circ} \times \frac{60}{10}$
$β = 18^{\circ}$

CG PET -2016

Ray Optics

282917 Distance between objective and eye- piece of a microscope is $20.6 cm$ . Consider both lens are thin and focal length of each lens is $6 m m$ . If last image is formed at infinity, then linear magnification of the objective is

1 -1347

2 -6.19

3 -32.3 times

4 -3.23 times

Explanation:

C: Given, $f_{o} = 6 mm, f_{e} = 6 mm$
Distance between them $(L) = 20.6 cm = 206 mm$ We know that-
$\begin{array}{r} L = v_{o} + f_{e} \\ 206 = v_{o} + 6 \\ v_{o} = 200 mm \end{array} {∵ v_{o} = image formed by objective}$
Using lens formula -
$\begin{array}{r} \frac{1}{f_{0}} = \frac{1}{v_{o}} - \frac{1}{u_{o}} \\ \frac{1}{6} = \frac{1}{200} - \frac{1}{u_{o}} \\ \frac{1}{u_{o}} = \frac{1}{200} - \frac{1}{6} \\ u_{o} = - 6.18 mm \end{array}$
Now, magnification, $m = \frac{- v}{u} = \frac{- 200}{6.18}$
$m = - 32.3 times$

CG PET- 2014

Ray Optics

282918 Magnifying power of an astronomical telescope will be maximum when the final image formed by it is

1 at infinity

2 at least distance of distinct vision

3 at any where

4 at optical centre of objective lens

CG PET- 2013

Ray Optics

282915 Two luminous point sources separated by a certain distance are at $10 km$ from an observer. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is $500 nm$ , 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:

C: Given,
$\begin{array}{r} λ_{1} = 500 nm = 500 \times 10^{- 9} m \\ d = 2.5 \times 10^{- 3} m \\ r = 10 km = 10, 000 m \end{array}$
According to Rayleigh's criterion,
Using
$\begin{array}{r} α = \frac{1.22 λ}{d} \\ α = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} \\ α = 2.44 \times 10^{- 4} \\ α = \frac{l}{r} \end{array}$
But,
$α = \frac{l}{r}$
Distance of separation $(l) = α . r$
$\begin{array}{r} l = 2.44 \times 10^{- 4} \times 10, 000 \\ l = 2.44 m \end{array}$

Manipal UGET-2009

Ray Optics

282916 In a telescope, the focal length of objective is 60 $cm$ and eye piece is $10 cm$ . When it is focused on an object parallel rays come out of eyepiece. If the angle subtended at objective by object is $3^{\circ}$ , then what is the angular width of image?

1

24^{\circ}

2

18^{\circ}

3

12^{\circ}

4

6^{\circ}

Explanation:

B: Given,
$f_{o} = 60 cm, f_{e} = 10 cm$
$α = 3^{\circ}$
We know that -
Magnification, $\frac{β}{α} = \frac{f_{o}}{f_{e}}$
Where, $β =$ angular width of image
$α =$ angle subtended at object
$f_{o} =$ focal length of objective
$f_{e} =$ focal length of eye piece
$β = 3^{\circ} \times \frac{60}{10}$
$β = 18^{\circ}$

CG PET -2016

Ray Optics

282917 Distance between objective and eye- piece of a microscope is $20.6 cm$ . Consider both lens are thin and focal length of each lens is $6 m m$ . If last image is formed at infinity, then linear magnification of the objective is

1 -1347

2 -6.19

3 -32.3 times

4 -3.23 times

Explanation:

C: Given, $f_{o} = 6 mm, f_{e} = 6 mm$
Distance between them $(L) = 20.6 cm = 206 mm$ We know that-
$\begin{array}{r} L = v_{o} + f_{e} \\ 206 = v_{o} + 6 \\ v_{o} = 200 mm \end{array} {∵ v_{o} = image formed by objective}$
Using lens formula -
$\begin{array}{r} \frac{1}{f_{0}} = \frac{1}{v_{o}} - \frac{1}{u_{o}} \\ \frac{1}{6} = \frac{1}{200} - \frac{1}{u_{o}} \\ \frac{1}{u_{o}} = \frac{1}{200} - \frac{1}{6} \\ u_{o} = - 6.18 mm \end{array}$
Now, magnification, $m = \frac{- v}{u} = \frac{- 200}{6.18}$
$m = - 32.3 times$

CG PET- 2014

Ray Optics

282918 Magnifying power of an astronomical telescope will be maximum when the final image formed by it is

1 at infinity

2 at least distance of distinct vision

3 at any where

4 at optical centre of objective lens

CG PET- 2013

Ray Optics

282915 Two luminous point sources separated by a certain distance are at $10 km$ from an observer. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is $500 nm$ , 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:

C: Given,
$\begin{array}{r} λ_{1} = 500 nm = 500 \times 10^{- 9} m \\ d = 2.5 \times 10^{- 3} m \\ r = 10 km = 10, 000 m \end{array}$
According to Rayleigh's criterion,
Using
$\begin{array}{r} α = \frac{1.22 λ}{d} \\ α = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} \\ α = 2.44 \times 10^{- 4} \\ α = \frac{l}{r} \end{array}$
But,
$α = \frac{l}{r}$
Distance of separation $(l) = α . r$
$\begin{array}{r} l = 2.44 \times 10^{- 4} \times 10, 000 \\ l = 2.44 m \end{array}$

Manipal UGET-2009

Ray Optics

282916 In a telescope, the focal length of objective is 60 $cm$ and eye piece is $10 cm$ . When it is focused on an object parallel rays come out of eyepiece. If the angle subtended at objective by object is $3^{\circ}$ , then what is the angular width of image?

1

24^{\circ}

2

18^{\circ}

3

12^{\circ}

4

6^{\circ}

Explanation:

B: Given,
$f_{o} = 60 cm, f_{e} = 10 cm$
$α = 3^{\circ}$
We know that -
Magnification, $\frac{β}{α} = \frac{f_{o}}{f_{e}}$
Where, $β =$ angular width of image
$α =$ angle subtended at object
$f_{o} =$ focal length of objective
$f_{e} =$ focal length of eye piece
$β = 3^{\circ} \times \frac{60}{10}$
$β = 18^{\circ}$

CG PET -2016

Ray Optics

282917 Distance between objective and eye- piece of a microscope is $20.6 cm$ . Consider both lens are thin and focal length of each lens is $6 m m$ . If last image is formed at infinity, then linear magnification of the objective is

1 -1347

2 -6.19

3 -32.3 times

4 -3.23 times

Explanation:

C: Given, $f_{o} = 6 mm, f_{e} = 6 mm$
Distance between them $(L) = 20.6 cm = 206 mm$ We know that-
$\begin{array}{r} L = v_{o} + f_{e} \\ 206 = v_{o} + 6 \\ v_{o} = 200 mm \end{array} {∵ v_{o} = image formed by objective}$
Using lens formula -
$\begin{array}{r} \frac{1}{f_{0}} = \frac{1}{v_{o}} - \frac{1}{u_{o}} \\ \frac{1}{6} = \frac{1}{200} - \frac{1}{u_{o}} \\ \frac{1}{u_{o}} = \frac{1}{200} - \frac{1}{6} \\ u_{o} = - 6.18 mm \end{array}$
Now, magnification, $m = \frac{- v}{u} = \frac{- 200}{6.18}$
$m = - 32.3 times$

CG PET- 2014

Ray Optics

282918 Magnifying power of an astronomical telescope will be maximum when the final image formed by it is

1 at infinity

2 at least distance of distinct vision

3 at any where

4 at optical centre of objective lens

CG PET- 2013

Ray Optics

282915 Two luminous point sources separated by a certain distance are at $10 km$ from an observer. If the aperture of his eye is $2.5 \times 10^{- 3} m$ and the wavelength of light used is $500 nm$ , 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:

C: Given,
$\begin{array}{r} λ_{1} = 500 nm = 500 \times 10^{- 9} m \\ d = 2.5 \times 10^{- 3} m \\ r = 10 km = 10, 000 m \end{array}$
According to Rayleigh's criterion,
Using
$\begin{array}{r} α = \frac{1.22 λ}{d} \\ α = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} \\ α = 2.44 \times 10^{- 4} \\ α = \frac{l}{r} \end{array}$
But,
$α = \frac{l}{r}$
Distance of separation $(l) = α . r$
$\begin{array}{r} l = 2.44 \times 10^{- 4} \times 10, 000 \\ l = 2.44 m \end{array}$

Manipal UGET-2009

Ray Optics

282916 In a telescope, the focal length of objective is 60 $cm$ and eye piece is $10 cm$ . When it is focused on an object parallel rays come out of eyepiece. If the angle subtended at objective by object is $3^{\circ}$ , then what is the angular width of image?

1

24^{\circ}

2

18^{\circ}

3

12^{\circ}

4

6^{\circ}

Explanation:

B: Given,
$f_{o} = 60 cm, f_{e} = 10 cm$
$α = 3^{\circ}$
We know that -
Magnification, $\frac{β}{α} = \frac{f_{o}}{f_{e}}$
Where, $β =$ angular width of image
$α =$ angle subtended at object
$f_{o} =$ focal length of objective
$f_{e} =$ focal length of eye piece
$β = 3^{\circ} \times \frac{60}{10}$
$β = 18^{\circ}$

CG PET -2016

Ray Optics

282917 Distance between objective and eye- piece of a microscope is $20.6 cm$ . Consider both lens are thin and focal length of each lens is $6 m m$ . If last image is formed at infinity, then linear magnification of the objective is

1 -1347

2 -6.19

3 -32.3 times

4 -3.23 times

Explanation:

C: Given, $f_{o} = 6 mm, f_{e} = 6 mm$
Distance between them $(L) = 20.6 cm = 206 mm$ We know that-
$\begin{array}{r} L = v_{o} + f_{e} \\ 206 = v_{o} + 6 \\ v_{o} = 200 mm \end{array} {∵ v_{o} = image formed by objective}$
Using lens formula -
$\begin{array}{r} \frac{1}{f_{0}} = \frac{1}{v_{o}} - \frac{1}{u_{o}} \\ \frac{1}{6} = \frac{1}{200} - \frac{1}{u_{o}} \\ \frac{1}{u_{o}} = \frac{1}{200} - \frac{1}{6} \\ u_{o} = - 6.18 mm \end{array}$
Now, magnification, $m = \frac{- v}{u} = \frac{- 200}{6.18}$
$m = - 32.3 times$

CG PET- 2014

Ray Optics

282918 Magnifying power of an astronomical telescope will be maximum when the final image formed by it is

1 at infinity

2 at least distance of distinct vision

3 at any where

4 at optical centre of objective lens

CG PET- 2013