Converging Lens Problem: Finding Distance from Slide for Real and Virtual Images

[nemzy]

A transparent photographic slide is placed in front of a converging lens with a focal length of 2.24 cm. The lens forms an image of the slide 11.4 cm from the slide.
(a) How far is the lens from the slide if the image is real?

(b) How far is the lens from the slide if the image is virtual?


I got the wrong answer, and i don't know why.

This is how i attempted part a:

a) focus=2.24 cm
q= 11.4 cm (positive, since the image is real)
1/p + 1/q = 1/f solve for p


b) i did same process except p = -11.4 cm (negative since virtual image)


But yet still got the wrong answer

What am i doing wrong? thanks

 

[ehild]

A transparent photographic slide is placed in front of a converging lens with a focal length of 2.24 cm. The lens forms an image of the slide 11.4 cm from the slide.
(a) How far is the lens from the slide if the image is real?

(b) How far is the lens from the slide if the image is virtual?


I got the wrong answer, and i don't know why.

This is how i attempted part a:

a) focus=2.24 cm
q= 11.4 cm (positive, since the image is real)
1/p + 1/q = 1/f solve for p

q is not 11.4 cm. q should be the distance of the slide from the lens, and 11.4 cm is said to be the distance between the slide and its image.
ehild

 

[wais]

It seems that you have the right approach for solving this problem. However, it is possible that you made a mistake in your calculations. Let's go through the steps together to see where the error might have occurred.

First, we need to identify the given values:

- Focal length (f) = 2.24 cm
- Image distance (q) = 11.4 cm
- Object distance (p) = unknown

Next, we can plug these values into the thin lens equation:

1/p + 1/q = 1/f

Solving for p, we get:

1/p = 1/f - 1/q

1/p = 1/2.24 - 1/11.4

1/p = 0.4464 - 0.0877

1/p = 0.3587

p = 1/0.3587

p = 2.79 cm

Therefore, the lens is 2.79 cm away from the slide if the image is real.

For part b, we can use the same equation but with a negative value for q since the image is virtual:

1/p = 1/f - 1/q

1/p = 1/2.24 - 1/-11.4

1/p = 0.4464 + 0.0877

1/p = 0.5341

p = 1/0.5341

p = 1.87 cm

So, the lens is 1.87 cm away from the slide if the image is virtual.

It's possible that you made a mistake in your calculations or accidentally used the wrong values. Double-check your work and see if you can find where the error occurred. Hopefully, this explanation helps you understand the problem better. Good luck!