QBManager

Ray Optics

282874 The magnifying power of a telescope is $m^{'}$ . If the focal length of the eye piece is doubled, then its magnifying power will become

1

\sqrt{2} m

2

3 m

3

2 m

4

\frac{m}{2}

Explanation:

D: For a telescope, Magnifying power
$m = \frac{focal length of objective (f_{0})}{focallength of eyepiece (f_{c})}$
$m = \frac{f_{0}}{f_{e}}$
If, $f_{s}^{'} = 2 f_{c}$
New magnifying power,
$m^{'} = \frac{f_{o}}{f_{e}^{'}} = \frac{f_{o}}{2 f_{e}} = \frac{m}{2}$

Manipal UGET -2020

Ray Optics

282875 Find the angular magnification of telescope if focal length of objective and eye lenses are 10 $cm$ and $10 mm$ respectively and tube length is $11 cm$ :

1 10

2 5

3 100

4 50

Explanation:

A: Given,
Focal length of object $(f_{o}) = 10 cm$ and focal length of eye lens $(f_{c}) = 10 mm = 1 cm$ As we know that,
Angular magnification of telescope $= \frac{f_{o}}{f_{e}} = \frac{10}{l} = 10$

AIIMS-26.05.2019(E) Shift-2

Ray Optics

282876 The limit of resolution of a telescope is $2.5 \times 10^{-}$ rad. If the telescope is used to detect light of wavelength $500 mm$ coming from a star, the diameter of the objective lens used by telescope is

1

244 cm

2

258 cm

3

228 cm

4

264 cm

Explanation:

A: Given,
Limit of resolution $(α) = 2.5 \times 10^{- 7}$ rad
Wavelength $(λ) = 500 nm = 500 \times 10^{- 9} m$
We know that,
Diameter of lens $(d) = \frac{1.22 λ}{α}$
$\begin{array}{r} d = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 7}} \\ d = 244 cm \end{array}$

TS-EAMCET-04.05.2019

Ray Optics

282877 The focal length of the objective lens of a telescope is $50 cm$ . If the magnification of the telescope is 25 , then the focal length of the eyepiece is

1

12.5 cm

2

5 cm

3

2 cm

4

10 cm

Explanation:

C: Given,
Focal length of objective $(f_{0}) = 50 cm$
Magnification $(m) = 25$
We know that.
Magnification of telescope,
$\begin{array}{r} m = \frac{f_{o}}{f_{e}} \\ 25 = \frac{50}{f_{e}} \\ f_{c} = \frac{50}{25} \\ f_{c} = 2 cm \end{array}$

NDA (I) 2019

Ray Optics

282874 The magnifying power of a telescope is $m^{'}$ . If the focal length of the eye piece is doubled, then its magnifying power will become

1

\sqrt{2} m

2

3 m

3

2 m

4

\frac{m}{2}

Explanation:

D: For a telescope, Magnifying power
$m = \frac{focal length of objective (f_{0})}{focallength of eyepiece (f_{c})}$
$m = \frac{f_{0}}{f_{e}}$
If, $f_{s}^{'} = 2 f_{c}$
New magnifying power,
$m^{'} = \frac{f_{o}}{f_{e}^{'}} = \frac{f_{o}}{2 f_{e}} = \frac{m}{2}$

Manipal UGET -2020

Ray Optics

282875 Find the angular magnification of telescope if focal length of objective and eye lenses are 10 $cm$ and $10 mm$ respectively and tube length is $11 cm$ :

1 10

2 5

3 100

4 50

Explanation:

A: Given,
Focal length of object $(f_{o}) = 10 cm$ and focal length of eye lens $(f_{c}) = 10 mm = 1 cm$ As we know that,
Angular magnification of telescope $= \frac{f_{o}}{f_{e}} = \frac{10}{l} = 10$

AIIMS-26.05.2019(E) Shift-2

Ray Optics

282876 The limit of resolution of a telescope is $2.5 \times 10^{-}$ rad. If the telescope is used to detect light of wavelength $500 mm$ coming from a star, the diameter of the objective lens used by telescope is

1

244 cm

2

258 cm

3

228 cm

4

264 cm

Explanation:

A: Given,
Limit of resolution $(α) = 2.5 \times 10^{- 7}$ rad
Wavelength $(λ) = 500 nm = 500 \times 10^{- 9} m$
We know that,
Diameter of lens $(d) = \frac{1.22 λ}{α}$
$\begin{array}{r} d = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 7}} \\ d = 244 cm \end{array}$

TS-EAMCET-04.05.2019

Ray Optics

282877 The focal length of the objective lens of a telescope is $50 cm$ . If the magnification of the telescope is 25 , then the focal length of the eyepiece is

1

12.5 cm

2

5 cm

3

2 cm

4

10 cm

Explanation:

C: Given,
Focal length of objective $(f_{0}) = 50 cm$
Magnification $(m) = 25$
We know that.
Magnification of telescope,
$\begin{array}{r} m = \frac{f_{o}}{f_{e}} \\ 25 = \frac{50}{f_{e}} \\ f_{c} = \frac{50}{25} \\ f_{c} = 2 cm \end{array}$

NDA (I) 2019

Ray Optics

282874 The magnifying power of a telescope is $m^{'}$ . If the focal length of the eye piece is doubled, then its magnifying power will become

1

\sqrt{2} m

2

3 m

3

2 m

4

\frac{m}{2}

Explanation:

D: For a telescope, Magnifying power
$m = \frac{focal length of objective (f_{0})}{focallength of eyepiece (f_{c})}$
$m = \frac{f_{0}}{f_{e}}$
If, $f_{s}^{'} = 2 f_{c}$
New magnifying power,
$m^{'} = \frac{f_{o}}{f_{e}^{'}} = \frac{f_{o}}{2 f_{e}} = \frac{m}{2}$

Manipal UGET -2020

Ray Optics

282875 Find the angular magnification of telescope if focal length of objective and eye lenses are 10 $cm$ and $10 mm$ respectively and tube length is $11 cm$ :

1 10

2 5

3 100

4 50

Explanation:

A: Given,
Focal length of object $(f_{o}) = 10 cm$ and focal length of eye lens $(f_{c}) = 10 mm = 1 cm$ As we know that,
Angular magnification of telescope $= \frac{f_{o}}{f_{e}} = \frac{10}{l} = 10$

AIIMS-26.05.2019(E) Shift-2

Ray Optics

282876 The limit of resolution of a telescope is $2.5 \times 10^{-}$ rad. If the telescope is used to detect light of wavelength $500 mm$ coming from a star, the diameter of the objective lens used by telescope is

1

244 cm

2

258 cm

3

228 cm

4

264 cm

Explanation:

A: Given,
Limit of resolution $(α) = 2.5 \times 10^{- 7}$ rad
Wavelength $(λ) = 500 nm = 500 \times 10^{- 9} m$
We know that,
Diameter of lens $(d) = \frac{1.22 λ}{α}$
$\begin{array}{r} d = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 7}} \\ d = 244 cm \end{array}$

TS-EAMCET-04.05.2019

Ray Optics

282877 The focal length of the objective lens of a telescope is $50 cm$ . If the magnification of the telescope is 25 , then the focal length of the eyepiece is

1

12.5 cm

2

5 cm

3

2 cm

4

10 cm

Explanation:

C: Given,
Focal length of objective $(f_{0}) = 50 cm$
Magnification $(m) = 25$
We know that.
Magnification of telescope,
$\begin{array}{r} m = \frac{f_{o}}{f_{e}} \\ 25 = \frac{50}{f_{e}} \\ f_{c} = \frac{50}{25} \\ f_{c} = 2 cm \end{array}$

NDA (I) 2019

Ray Optics

282874 The magnifying power of a telescope is $m^{'}$ . If the focal length of the eye piece is doubled, then its magnifying power will become

1

\sqrt{2} m

2

3 m

3

2 m

4

\frac{m}{2}

Explanation:

D: For a telescope, Magnifying power
$m = \frac{focal length of objective (f_{0})}{focallength of eyepiece (f_{c})}$
$m = \frac{f_{0}}{f_{e}}$
If, $f_{s}^{'} = 2 f_{c}$
New magnifying power,
$m^{'} = \frac{f_{o}}{f_{e}^{'}} = \frac{f_{o}}{2 f_{e}} = \frac{m}{2}$

Manipal UGET -2020

Ray Optics

282875 Find the angular magnification of telescope if focal length of objective and eye lenses are 10 $cm$ and $10 mm$ respectively and tube length is $11 cm$ :

1 10

2 5

3 100

4 50

Explanation:

A: Given,
Focal length of object $(f_{o}) = 10 cm$ and focal length of eye lens $(f_{c}) = 10 mm = 1 cm$ As we know that,
Angular magnification of telescope $= \frac{f_{o}}{f_{e}} = \frac{10}{l} = 10$

AIIMS-26.05.2019(E) Shift-2

Ray Optics

282876 The limit of resolution of a telescope is $2.5 \times 10^{-}$ rad. If the telescope is used to detect light of wavelength $500 mm$ coming from a star, the diameter of the objective lens used by telescope is

1

244 cm

2

258 cm

3

228 cm

4

264 cm

Explanation:

A: Given,
Limit of resolution $(α) = 2.5 \times 10^{- 7}$ rad
Wavelength $(λ) = 500 nm = 500 \times 10^{- 9} m$
We know that,
Diameter of lens $(d) = \frac{1.22 λ}{α}$
$\begin{array}{r} d = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 7}} \\ d = 244 cm \end{array}$

TS-EAMCET-04.05.2019

Ray Optics

282877 The focal length of the objective lens of a telescope is $50 cm$ . If the magnification of the telescope is 25 , then the focal length of the eyepiece is

1

12.5 cm

2

5 cm

3

2 cm

4

10 cm

Explanation:

C: Given,
Focal length of objective $(f_{0}) = 50 cm$
Magnification $(m) = 25$
We know that.
Magnification of telescope,
$\begin{array}{r} m = \frac{f_{o}}{f_{e}} \\ 25 = \frac{50}{f_{e}} \\ f_{c} = \frac{50}{25} \\ f_{c} = 2 cm \end{array}$

NDA (I) 2019