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kaheart [24]
3 years ago
15

A doctor examines a mole with a 15.0 cm focal length magnifying glass held 13.5 cm from the mole. Where is the image? what is it

s magnification? how big is the image of a 5.00 mm diameter mole?
Physics
1 answer:
daser333 [38]3 years ago
4 0

1. Virtual, 135 from the lens

The distance of the image from the lens can be calculated by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the mole from the lens

In this problem, we have

f = 15.0 cm is the focal length (positive for a convex lens, as the one in a magnifying glass)

p = 13.5 cm is the distance of the mole from the lens

Solving the equation for q,

q=\frac{1}{\frac{1}{15.0 cm}-\frac{1}{13.5 cm}}=-135 cm

The negative sign tells that the image is virtual (on the opposite side of the lens with respect to the image), and located 135 cm from the lens.

2. 10

The magnification M of the image is given by

M=-\frac{q}{p}

where

q = -135 cm is the distance of the image from the lens

p = 13.5 cm is the distance of the mole from the lens

Solving the equation for M, we find

M=-\frac{-135 cm}{13.5 cm}=10

3. 50 mm

The magnification equation can also be written as

M=\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the size of the image

h_o is the size of the object

Since here we have

h_o = 5.00 mm diameter of the real mole

M = 10

We find

h_i = M h_o = (10)(5.00 mm)=50 mm

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Where:

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V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

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V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

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