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3 years ago

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We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu

st be the focal length of the appropriate positive lens? If it is a bi-convex lens of refractive index 1.5, what are the values of radii of curvature?

1 answer:

liubo4ka [24]3 years ago

3 0

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

$\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}$

$\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}$

f = 30 cm

using lens formula

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})$

$R_1 = R\ and\ R_2 = -R$

$\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})$

$R = (n -1)\ f$

$R = 2(1.5 -1)\ 30$

R = 30 cm

hence, the radii of curvature is 30 cm.

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Answer:

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$\mathrm{v}_{2}=\frac{3000}{900}$

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4 years ago

On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow

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Answer:

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4 years ago

The gravitational potential energy of a particle of mass m moving under the influence of a fixed mass M is given by - , where G

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A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.

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1 year ago

Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

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Explanation:

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coefficient of static friction between the object and the surface, $\mu_s=0.1$

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(b)

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$U=F_k.d$

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$d=0.01153\ m=11.53\ mm$

<em>is the total distance does it travel before stopping.</em>

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3 years ago