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1. what is the magnificant of a telescope if the focal lengths of its objective and its eyepiece are 200 cm. and 2 cm. respectively?

2. a converging lens with an 18 cm. focal length is used for a lens of a projector. if a film is placed 20 cm. away from the lens, how far must the screen be placed from the lens? ​

Sagot :

Mizuidn

Lens Equation

To find the distance of the image from the lens, we use the formula:

[tex]\frac{1}{f} =\frac{1}{p}+\frac{1}{q}[/tex]

where:

f = focal length

p = distance of the object from the lens

q = distance of the image from the lens

Answers

1.

Given

f = 200 cm

p = 2 cm

q = ?

Solve

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q} \\ \frac{1}{200\:cm} =\frac{1}{2\:cm}+\frac{1}{q} \\ \frac{1}{200\:cm} -\frac{1}{2\:cm}=\frac{1}{q}\\ \frac{2\:cm\:-\:200\:cm}{(200\:cm)\:(2\:cm)} =\frac{1}{q} \\ \frac{-198\:cm}{400\:cm}=\frac{1}{q} \\ q=-2\:cm[/tex]

∴ The magnification of the telescope is -2 cm.

2.

Given

f = 18 cm

p = 20 cm

q = ?

Solve

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q} \\ \frac{1}{18\:cm}=\frac{1}{20\:cm}+\frac{1}{q} \\ \frac{1}{18\:cm}-\frac{1}{20\:cm}=\frac{1}{q}\\ \frac{20\:cm\:-\:18\:cm}{(18\:cm)(20\:cm)} =\frac{1}{q} \\ \frac{2\:cm}{360\:cm}=\frac{1}{q} \\ q=180\:cm[/tex]

∴ The screen is 180 cm away from the lens.

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