Answer:
a. Magnification = 6.1
b. The image formed is virtual, and on the same side of the lens as the object.
c. Image size = 119.8 squared millimetres
Explanation:
Magnification = [tex]\frac{Image distance}{Object distance}[/tex]
But, focal length, f = 15.0 cm, and object distance, u = 12.4 cm. Let the image distance be represented by v.
a. Applying the lens formula, we have;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{15}[/tex] = [tex]\frac{1}{12.4}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{15}[/tex] - [tex]\frac{1}{12.4}[/tex]
= -[tex]\frac{13}{930}[/tex]
v = -75.1538
The image distance, v = -75.2 cm
Magnification = [tex]\frac{75.2}{12.4}[/tex]
= 6.0645
Magnification = 6.1
b. The image formed is virtual, and on the same side of the lens as the object.
c. Given that diameter of mole = 5.00 mm.
Its radius = [tex]\frac{diameter}{2}[/tex] = [tex]\frac{5.0}{2}[/tex]
= 2.5 mm
Thus, the area of the mole would be;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](2.5)^{2}[/tex]
= 19.643
A = 19.64 square millimetres.
Thus, the size of the image can be determined by;
Magnification = [tex]\frac{Image size}{Object size}[/tex]
Image size = Magnification x object size
= 6.1 x 19.64
= 119.804
The size of the image is 119.8 squared millimetres.
