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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364543 A small telescope has an objective lens of focal length $140 c m$ and eye piece of focal length $5.0 c m$ . The telescope is used to view a $100 m$ tall tower $3 k m$ away. The height of the image of the tower formed by objective lens is

1

\frac{14}{3} c m

2

\frac{11}{3} c m

3

\frac{17}{3} c m

4

\frac{8}{3} c m

Explanation:

The height of the image of the tower is given by
$m = \frac{f_{0}}{u_{0}} = \frac{h_{2}}{h_{1}} \Rightarrow h_{2} = \frac{f_{0} h_{1}}{u_{0}} = \frac{14}{3} c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364544 An astronomical telescope with magnification 50 is to be designed in normal adjustment. Length of the tube is $102 c m$ . The powers of objective and eyepiece are respectively.

1 2D, 50 D

2 1.5D, 20 D

3 1D, 40 D

4 1D, 50 D

Explanation:

$m = \frac{f_{0}}{f_{e}}$ and $L = f_{0} + f_{e} \Rightarrow \frac{f_{0}}{f_{e}} = 50$
$f_{0} + f_{e} = 102$
$\Rightarrow f_{0} + \frac{f_{0}}{50} = 102$ $\Rightarrow f_{0} = 100$
$\Rightarrow f_{e} = 2$
$P_{e} = \frac{100}{2} = 50 D, P_{0} = \frac{1}{1 m} = 1 m^{- 1} = 1 D$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364545 An astronomical telescope has objective and eyepiece lens of powers $0.5 D$ and $20 D$ respectively, its magnifying power will be

1 8

2 20

3 30

4 40

Explanation:

$m = \frac{f_{o}}{f_{e}} = \frac{P_{e}}{P_{o}} = \frac{20}{0.5} = 40.$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364546 A giant telescope in an observatory has objective of focal length $19 c m$ and an eyepiece of focal length $1.0 c m$ . In normal adjustment, the telescope is used to view the moon. Diameter of image of the moon formed by the objective is (The diameter of the moon is $3.5 \times 10^{6} m$ and the radius of the lunar orbit round the earth is $3.8 \times 10^{8} m$ )

1

10 c m

2

12.5 c m

3

15 c m

4

17.5 c m

Explanation:

The diameter of the image formed by the objective lens is
$\tan θ = \frac{D_{1} / 2}{| u_{0} |} = \frac{D_{1} / 2}{f_{0}}$
supporting img
$D_{2} = \frac{f_{0}}{| u_{0} |} D_{1} = \frac{19 m \times 3.5 \times 10^{6}}{3.8 \times 10^{8}} = 17.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364543 A small telescope has an objective lens of focal length $140 c m$ and eye piece of focal length $5.0 c m$ . The telescope is used to view a $100 m$ tall tower $3 k m$ away. The height of the image of the tower formed by objective lens is

1

\frac{14}{3} c m

2

\frac{11}{3} c m

3

\frac{17}{3} c m

4

\frac{8}{3} c m

Explanation:

The height of the image of the tower is given by
$m = \frac{f_{0}}{u_{0}} = \frac{h_{2}}{h_{1}} \Rightarrow h_{2} = \frac{f_{0} h_{1}}{u_{0}} = \frac{14}{3} c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364544 An astronomical telescope with magnification 50 is to be designed in normal adjustment. Length of the tube is $102 c m$ . The powers of objective and eyepiece are respectively.

1 2D, 50 D

2 1.5D, 20 D

3 1D, 40 D

4 1D, 50 D

Explanation:

$m = \frac{f_{0}}{f_{e}}$ and $L = f_{0} + f_{e} \Rightarrow \frac{f_{0}}{f_{e}} = 50$
$f_{0} + f_{e} = 102$
$\Rightarrow f_{0} + \frac{f_{0}}{50} = 102$ $\Rightarrow f_{0} = 100$
$\Rightarrow f_{e} = 2$
$P_{e} = \frac{100}{2} = 50 D, P_{0} = \frac{1}{1 m} = 1 m^{- 1} = 1 D$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364545 An astronomical telescope has objective and eyepiece lens of powers $0.5 D$ and $20 D$ respectively, its magnifying power will be

1 8

2 20

3 30

4 40

Explanation:

$m = \frac{f_{o}}{f_{e}} = \frac{P_{e}}{P_{o}} = \frac{20}{0.5} = 40.$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364546 A giant telescope in an observatory has objective of focal length $19 c m$ and an eyepiece of focal length $1.0 c m$ . In normal adjustment, the telescope is used to view the moon. Diameter of image of the moon formed by the objective is (The diameter of the moon is $3.5 \times 10^{6} m$ and the radius of the lunar orbit round the earth is $3.8 \times 10^{8} m$ )

1

10 c m

2

12.5 c m

3

15 c m

4

17.5 c m

Explanation:

The diameter of the image formed by the objective lens is
$\tan θ = \frac{D_{1} / 2}{| u_{0} |} = \frac{D_{1} / 2}{f_{0}}$
supporting img
$D_{2} = \frac{f_{0}}{| u_{0} |} D_{1} = \frac{19 m \times 3.5 \times 10^{6}}{3.8 \times 10^{8}} = 17.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364543 A small telescope has an objective lens of focal length $140 c m$ and eye piece of focal length $5.0 c m$ . The telescope is used to view a $100 m$ tall tower $3 k m$ away. The height of the image of the tower formed by objective lens is

1

\frac{14}{3} c m

2

\frac{11}{3} c m

3

\frac{17}{3} c m

4

\frac{8}{3} c m

Explanation:

The height of the image of the tower is given by
$m = \frac{f_{0}}{u_{0}} = \frac{h_{2}}{h_{1}} \Rightarrow h_{2} = \frac{f_{0} h_{1}}{u_{0}} = \frac{14}{3} c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364544 An astronomical telescope with magnification 50 is to be designed in normal adjustment. Length of the tube is $102 c m$ . The powers of objective and eyepiece are respectively.

1 2D, 50 D

2 1.5D, 20 D

3 1D, 40 D

4 1D, 50 D

Explanation:

$m = \frac{f_{0}}{f_{e}}$ and $L = f_{0} + f_{e} \Rightarrow \frac{f_{0}}{f_{e}} = 50$
$f_{0} + f_{e} = 102$
$\Rightarrow f_{0} + \frac{f_{0}}{50} = 102$ $\Rightarrow f_{0} = 100$
$\Rightarrow f_{e} = 2$
$P_{e} = \frac{100}{2} = 50 D, P_{0} = \frac{1}{1 m} = 1 m^{- 1} = 1 D$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364545 An astronomical telescope has objective and eyepiece lens of powers $0.5 D$ and $20 D$ respectively, its magnifying power will be

1 8

2 20

3 30

4 40

Explanation:

$m = \frac{f_{o}}{f_{e}} = \frac{P_{e}}{P_{o}} = \frac{20}{0.5} = 40.$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364546 A giant telescope in an observatory has objective of focal length $19 c m$ and an eyepiece of focal length $1.0 c m$ . In normal adjustment, the telescope is used to view the moon. Diameter of image of the moon formed by the objective is (The diameter of the moon is $3.5 \times 10^{6} m$ and the radius of the lunar orbit round the earth is $3.8 \times 10^{8} m$ )

1

10 c m

2

12.5 c m

3

15 c m

4

17.5 c m

Explanation:

The diameter of the image formed by the objective lens is
$\tan θ = \frac{D_{1} / 2}{| u_{0} |} = \frac{D_{1} / 2}{f_{0}}$
supporting img
$D_{2} = \frac{f_{0}}{| u_{0} |} D_{1} = \frac{19 m \times 3.5 \times 10^{6}}{3.8 \times 10^{8}} = 17.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364543 A small telescope has an objective lens of focal length $140 c m$ and eye piece of focal length $5.0 c m$ . The telescope is used to view a $100 m$ tall tower $3 k m$ away. The height of the image of the tower formed by objective lens is

1

\frac{14}{3} c m

2

\frac{11}{3} c m

3

\frac{17}{3} c m

4

\frac{8}{3} c m

Explanation:

The height of the image of the tower is given by
$m = \frac{f_{0}}{u_{0}} = \frac{h_{2}}{h_{1}} \Rightarrow h_{2} = \frac{f_{0} h_{1}}{u_{0}} = \frac{14}{3} c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364544 An astronomical telescope with magnification 50 is to be designed in normal adjustment. Length of the tube is $102 c m$ . The powers of objective and eyepiece are respectively.

1 2D, 50 D

2 1.5D, 20 D

3 1D, 40 D

4 1D, 50 D

Explanation:

$m = \frac{f_{0}}{f_{e}}$ and $L = f_{0} + f_{e} \Rightarrow \frac{f_{0}}{f_{e}} = 50$
$f_{0} + f_{e} = 102$
$\Rightarrow f_{0} + \frac{f_{0}}{50} = 102$ $\Rightarrow f_{0} = 100$
$\Rightarrow f_{e} = 2$
$P_{e} = \frac{100}{2} = 50 D, P_{0} = \frac{1}{1 m} = 1 m^{- 1} = 1 D$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364545 An astronomical telescope has objective and eyepiece lens of powers $0.5 D$ and $20 D$ respectively, its magnifying power will be

1 8

2 20

3 30

4 40

Explanation:

$m = \frac{f_{o}}{f_{e}} = \frac{P_{e}}{P_{o}} = \frac{20}{0.5} = 40.$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364546 A giant telescope in an observatory has objective of focal length $19 c m$ and an eyepiece of focal length $1.0 c m$ . In normal adjustment, the telescope is used to view the moon. Diameter of image of the moon formed by the objective is (The diameter of the moon is $3.5 \times 10^{6} m$ and the radius of the lunar orbit round the earth is $3.8 \times 10^{8} m$ )

1

10 c m

2

12.5 c m

3

15 c m

4

17.5 c m

Explanation:

The diameter of the image formed by the objective lens is
$\tan θ = \frac{D_{1} / 2}{| u_{0} |} = \frac{D_{1} / 2}{f_{0}}$
supporting img
$D_{2} = \frac{f_{0}}{| u_{0} |} D_{1} = \frac{19 m \times 3.5 \times 10^{6}}{3.8 \times 10^{8}} = 17.5 c m$

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