1. Virtual, 135 from the lens
The distance of the image from the lens can be calculated by using the lens equation:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}[/tex]
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,
[tex]q=\frac{1}{\frac{1}{15.0 cm}-\frac{1}{13.5 cm}}=-135 cm[/tex]
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
[tex]M=-\frac{q}{p}[/tex]
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
[tex]M=-\frac{-135 cm}{13.5 cm}=10[/tex]
3. 50 mm
The magnification equation can also be written as
[tex]M=\frac{h_i}{h_o}=-\frac{q}{p}[/tex]
where
[tex]h_i[/tex] is the size of the image
[tex]h_o[/tex] is the size of the object
Since here we have
[tex]h_o = 5.00 mm[/tex] diameter of the real mole
M = 10
We find
[tex]h_i = M h_o = (10)(5.00 mm)=50 mm[/tex]
(a) The position of the image is -135.1 cm.
(b) magnification of the lens is 10
(c) The size of image formed by 5 mm diameter mole is 50 mm.
The image distance is determined by applying lens formula as hwon below;
1/f = 1/v + 1/u
1/v = 1/f - 1/u
1/v = 1/15 - 1/13.5
1/v = -0.0074
v = -135.1 cm
The magnification of the lens is calcuated as follows;
M = v/u
M = 135.1/13.5
M = 10
M = 10 x 5 mm
M = 50 mm
Learn more about magnification of lens here: https://brainly.com/question/25779311
