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Elenna [48]
3 years ago
15

The objective lens in a microscope with a 17.0 cm long tube has a magnification of -50.0 and the eyepiece has a magnification of

20.0. (a) What is the focal length of the objective? (b) What is the focal length of the eyepiece?
Physics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

(a) 0.34 cm

(b) 1.32 cm

Explanation:

Given that, m_{o}=-50

m_{e}=20 cm

And the length of objective lens in a microscope

L=17

And we have d of the eyepiece is 25 cm.

(a) Focal length of the objective is,

f_{o}=-\frac{L}{m_{o} }\\  f_{o}=-\frac{17}{-50} \\f_{o}=0.34 cm

Therefore, objective lens's focal length is 0.34 cm.

(b) Focal length of an eyepiece,

m_{e}=(1+\frac{d}{f_{e} })\\  20=(1+\frac{25}{f_{e} }\\ f_{e}=\frac{25}{19}cm\\ f_{e}=1.32cm

Therefore, eyepiece's focal length is 1.32 cm.

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Answer:

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Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

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P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

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W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

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