Physics - Please Help with Question 2

Question

Please Help with Question 2 and can you double check my work 1 a-e. didn’t quite understand this lesson so please help me draw the diagrams 🙏🙏 even if you just draw one and explain how to do it in simple terms i would be so grateful

1. Find the distance to the image in the following lens problems and state whether the image is virtual or real AND whether it is upright or inverted. (4 points each)

a. An object is placed 40. mm away from a diverging lens with a focal length of 10. cm.

The distance to the image is -12.5 cm. The image is virtual and upright. 

 

b. An object is placed 2.5 cm away from a converging lens with a focal length of 1.5 cm.

The distance to the image is 2.1 cm. The image is real and inverted.

 

c. An object is placed 7.0 cm away from a converging lens with a focal length of 7.0 cm.

The distance to the image is 8 cm. The image is real and inverted. 

 

d. An object is placed 65 mm away from a diverging lens with a focal length of 30. mm.

The distance to the image is 35 cm. The image is real and upright.

 

e. An object is placed 12 cm away from a diverging lens with a focal length of 6.0 cm.

The distance to the image is -10 cm. The image is virtual and upright. 

2. Choose three of the above lens problems and draw a scale diagram for each one that shows geometrically how the distance to the image is found. (5 points each)

 

Be sure to: 

• Label so, si, and f clearly.

• Use a ruler or straight edge for neatness

• Draw to scale.

 

 

Answer 1

It can be helpful to rewrite the thin lens equation to solve for p directly in some of these:

1/s + 1/p = 1/f

1/p = 1/f - 1/s

p = 1 / (1/f - 1/s). Multiply the top and bottom by sf gives:

p = sf / (sf/f - sf/s)

p = sf / (s-f)

1a) Correct!

1b) Given s = 2.5 cm and f = +1.5 cm (since it converges), from p=sf/(s-f), I'm getting p = +3.75 cm, which means it's real and inverted

1c) Since the object distance is the same as the focal length (which is positive for converging lenses), the image distance must be at infinity: 1/s + 1/p = 1/f. If s = f, p = infinity

1d) Given s = 65 mm and f = -30 mm (since it diverges), using p = sf/(s-f), I'm getting p = -20.52 mm, which means it's virtual and upright

1e) Given s = 12 cm and f = -6 cm (since it diverges), using p = sf/(s-f), I'm getting p = -4 cm, which means it's virtual and upright

Ray tracing is a real pain! Starting with the object, draw a parallel line at some random height. Once that line hits the lens, draw a second line from that point to the focal length (behind the lens for a +f, in front of the lens for a -f). Next, draw a line from your original point through the center of the lens, onto the other side. Where those 2nd and 3rd lines intersect is where the image forms!

Below are links to some examples. First for a convex, converging lens.

https://image1.slideserve.com/2123350/ray-diagrams-l.jpg

Next, for a concave, diverging lens

https://www.leydenscience.org/physics/electmag/cavelens.jpg