Question

A 3.0-cm-tall object is 50 cm in front of a converging lens that has a 25 cm focal length. You may want to review (Pages 589 - 592) . Part A Calculate the image position. Express your answer to two significant figures and include the appropriate units. s′ s ′ = nothing nothing Request Answer Part B Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

Answer 1

A) 50 cm from the lens

We can use the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where

f is the focal length

s is the object distance

s' is the image distance

In this problem, we have

f = 25 cm

s = 50 cm

So using the formula we can find the image distance, s':

$\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}=\frac{1}{25 cm}-\frac{1}{50 cm}=\frac{1}{50 cm}\\s' = 50 cm$

and the positive sign means the image is real.

B) -3.0 cm (inverted image)

The image height is given by the magnification equation:

$\frac{y'}{y}=-\frac{s'}{s}$

where

y' is the image height

y = 3.0 cm is the object height

s' = 50 cm is the image distance

s = 50 cm is the object distance

Substituting into the formula, we find

$y'=-y\frac{s'}{s}=-(3.0 cm)\frac{50 cm}{50 cm}=-3.0 cm$

And the negative sign means the image is inverted.

Answer 2

The image position is 50 cm and the image height is 3 cm.

The image is same size as the object, real and inverted.

Image distance

The image distance is calculated using the following lens equations;

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

where;

• f is the focal length
• v is the image distance
• u is the object distance

$\frac{1}{v} = \frac{1}{25} - \frac{1}{50} \\\\\frac{1}{v} = \frac{2-1}{50} \\\\v = 50 \ cm$

Magnification of the image

$m = \frac{v}{u} \\\\m = \frac{50}{50} =1$

Thus, the image is same size as the object, real and inverted.

Learn more about converging lens here: https://brainly.com/question/24305458