Sagot :
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.