A doctor examines a mole with a 15.0 cm focal length magnifying glass held 12.4 cm from the mole. a. What is its magnification?
b. Where is the image?
c. How big is the image of a 5.00 mm diameter mole?

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 =

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;

= +

= +

= -

  = -

v = -75.1538

The image distance, v = -75.2 cm

Magnification =

                       = 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 = =

              = 2.5 mm

Thus, the area of the mole would be;

A =

   = x

   = 19.643

A = 19.64 square millimetres.

Thus, the size of the image can be determined by;

Magnification =

Image size = Magnification x object size

                  = 6.1 x 19.64

                  = 119.804

The size of the image is 119.8 squared millimetres.

c

explanation:

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.