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2 years ago

15

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

yepiece whose focal length is 2.7 cm

1 answer:

notka56 [123]2 years ago

4 0

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

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Tresset [83]

Answer:

Option A. 39.2 m/s

Explanation:

From the question given above, the following data were obtained:

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Since the initial velocity (u) is 0, the above equation becomes:

v = gt

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v = 9.8 × 4

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ss7ja [257]

Answer:

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Explanation:

Given that,

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mezya [45]

Answer:

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Explanation:

Given that,

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Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

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Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

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$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

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Using formula of angle

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