Question

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the image formed by the lens?
It is virtual, inverted, and smaller than the object.
It is real, inverted, and larger than the object.
It is real, upright, and larger than the object.
It is virtual, upright, and larger than the object.
It is real, inverted, and smaller than the object.

Answer 1

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

$\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}$

$q=\frac{1}{0.047 cm^{-1}}=+21.3 cm$

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

$h_i = - h_o \frac{q}{p}$

where  

$h_i, h_o$ are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

$h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o$

So we notice that:

$|h_i| < |h_o|$ : this means that the image is smaller than the object

$h_i < 0$ : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.