Question

A thin, convergent lens has a focal length of 8.00 cm. if a real, inverted image is located 12.0 cm to the right of the lens, where is the object located?

A. 24.0 cm to the right of the lens
B. 12.0 cm to the left of the lens
C. 24.0 cm to the left of the lens
D. 18.0 cm to the left of the lens
E. 18.0 cm, to the right of the lens

Answer 1

The object is located at 24.0 cm to the left of the lens using the lens formula which relates the focal length, object distance, and image distance for a thin converging lens.

Step-by-step explanation:

To locate the position of the object when given the image distance and the focal length of a convergent lens, we use the lens formula:

\( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

Where:

• f is the focal length of the lens,
• do is the object distance from the lens,
• di is the image distance from the lens (positive for real images, negative for virtual images).

In this problem, we have:

• f = 8.00 cm (since the lens is converging, f is positive),
• di = 12.0 cm (the image is real and on the same side as the object),

Substitute f and di into the lens formula:

\( \frac{1}{8.00} = \frac{1}{d_o} + \frac{1}{12.0} \)

\( \frac{1}{d_o} = \frac{1}{8.00} - \frac{1}{12.0} \)

\( \frac{1}{d_o} = \frac{3}{24.0} - \frac{2}{24.0} \)

\( \frac{1}{d_o} = \frac{1}{24.0} \)

Now calculate do:

\( d_o = 24.0 \ cm \)

The object is located 24.0 cm to the left of the lens. Therefore, the correct answer is:

C. 24.0 cm to the left of the lens

Answer 2

Using the lens formula with the given focal length of 8.00 cm and the real image distance of 12.0 cm, we find that the object is located 24.0 cm to the left of the lens, corresponding to option C.

Step-by-step explanation:

To determine where the object is located when a real, inverted image is formed by a thin convergent lens, we use the lens formula, which relates the object distance (do), the image distance (di), and the focal length (f):

\( \frac{1}{f} = \frac{1}{do} + \frac{1}{di} \)

We are given that the focal length, f, is 8.00 cm, and the real image is located 12.0 cm to the right of the lens (which means for a real image, di is positive and equals 12.0 cm). Plugging the given values into the lens formula:

\( \frac{1}{8.0} = \frac{1}{do} + \frac{1}{12.0} \)

Solving this equation for do gives us:

\( \frac{1}{do} = \frac{1}{8.0} - \frac{1}{12.0} \)

\( \frac{1}{do} = \frac{3 - 2}{24.0} \)

\( \frac{1}{do} = \frac{1}{24.0} \)

\( do = 24.0 \ cm \)

Therefore, the object is located 24.0 cm to the left of the lens, which matches option C. The sign convention for lens formulas implies that a positive object distance corresponds to the object being on the same side of the lens from where the light is coming.