An object 20cm high is placed 45cm from the lens of focal length 15cm? calculate the image distance and the size of the image and the nature of the image? If a concvex lens and a concave lens is used.

Question

An object 20cm high is placed 45cm from the lens of focal length 15cm? calculate the image distance and the  size of the image and the nature of the image? If a concvex lens and a concave lens is used.

Julia Aguilar · Accepted Answer

Image distance #d_i = 22.5 " cm"#

The image size is #h_i = -10 " cm"#, meaning that the image is inverted.

Given: #h_o = 20 " cm"# #d_o = 45 " cm"# To calculate image distance, use the thin lens equation: #(1/d_o) + (1/d_i) = 1/f# #(1/45) + (1/d_i) = 1/15# #1/(d_i) = 2/45# To calculate image size, use the magnification equation: #m = frac{h_i}{h_o} = frac{-d_i}{d_o}# #frac{h_i}{20} = frac{-22.5}{45} #h_i = (20)*(-1/2)# If a convex lens is used, I think the image would be real. However, if a concave lens were used, I think the image would be virtual because it is a diverging lens.

Axel Acevedo · Answer

undefined Using the lens formula, $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $, where $ f $ is the focal length of the lens, $ v $ is the image distance, and $ u $ is the object distance:

Given:
Object distance ($ u $) = 45 cm
Focal length ($ f $) of convex lens = 15 cm

Calculating the image distance ($ v $):
$$ \frac{1}{15} = \frac{1}{v} + \frac{1}{45} $$
$$ \frac{1}{v} = \frac{1}{15} - \frac{1}{45} $$
$$ \frac{1}{v} = \frac{3}{45} - \frac{1}{45} $$
$$ \frac{1}{v} = \frac{2}{45} $$
$$ v = \frac{45}{2} $$
$$ v = 22.5 \text{ cm} $$

Now, to find the magnification ($ M $):
$$ M = \frac{v}{u} $$
$$ M = \frac{22.5}{45} $$
$$ M = \frac{1}{2} $$

Since $ M = \frac{1}{2} $, the image is half the size of the object.

To determine the nature of the image, we use the magnification:
- $ M > 0 $ for erect image
- $ M < 0 $ for inverted image

Since $ M = \frac{1}{2} $, which is positive, the image formed by the convex lens is erect.

If a concave lens is used, the calculation process is the same. However, for a concave lens, the focal length is negative. So, when plugging the values into the lens formula, the sign convention must be observed. The rest of the calculations proceed as before.