Calculating the Effects of Lens Closeness on Light Refraction

In summary, the conversation discusses the initial calculation of the position of the image formed by lens B when it is placed 25cm to the right of lens A. It is determined that the image is formed at infinity, with the rays emerging parallel. The discussion then moves on to determining the position of the image when lens B is moved closer to lens A. It is concluded that the image formed by lens B will always be negative and the rays will continue to emerge in a parallel direction. The use of a ray-diagram with a virtual object can also aid in understanding the positioning and direction of rays in this scenario.
  • #1
hidemi
208
36
Homework Statement
The two lenses shown are illuminated by a beam of parallel light from the left. Lens B is then moved slowly toward lens A. The beam emerging from lens B is:
A. initially parallel and then diverging
B. always diverging
C. initially converging and finally parallel
D. always parallel E. initially converging and finally diverging

The answer is A
Relevant Equations
1/f = 1/f1 + 1/f2 - x/f1*f2
I know what happens initially, calculated as follows.

1/∞ + 1/q = 1/50, q=50

1/-25 + 1/q = 1/-25, q= ∞

However, how do we know about the after when the two lenses get closer to each other?
 

Attachments

  • 1.png
    1.png
    12.1 KB · Views: 103
Physics news on Phys.org
  • #2
hidemi said:
I know what happens initially, calculated as follows.

1/∞ + 1/q = 1/50, q=50

1/-25 + 1/q = 1/-25, q= ∞

However, how do we know about the after when the two lenses get closer to each other?
Hi. Well no one has answered yet so see if this helps.

Lens A alone forms an image 50cm to its right.

Lens A’s image acts as a virtual object for lens B, initially 25cm to the right of lens B.

As you have done, you can then calculate the initial position of lens B’s image. It is infinity – so the rays from lens B emerge parallel.

If you can figure out how to draw a ray-diagram for lens B with a virtual object, you can do it without any calculations. (Note, lens B’s virtual object is at a focal point of lens B!)

If lens B is now moved left, closer to lens A, then lens B’s virtual object is now further to its right.

If ##u, v## and ##f## are the object-disance, image-distance and focal length of lens B then:

##\frac 1 u + \frac 1 v = \frac 1 f##

##v = \frac {uf}{u – f}##

In this question the numerator of the above equation is always positive because ##u## and ##f## are both negative. The denominator is negative when ##u <f##, i.e. when u is more negative than -25cm.

This means when lens B is moved (‘slowly’ or not!) towards lens A, v is always negative.

What does this tell you about the image formed by lens B? And hence what does this tell you about the direction of the rays from B?

Of course, using a ray-diagram for lens B with a virtual object, you could, do this with no calculations!
 
  • Like
Likes hidemi
  • #3
I got it. Thanks for helping.
 

Related to Calculating the Effects of Lens Closeness on Light Refraction

1. How does lens closeness affect light refraction?

Lens closeness refers to the distance between the lens and the object being viewed. The closer the lens is to the object, the greater the refraction of light will be. This is because the light rays have a shorter distance to travel through the lens, causing them to bend more.

2. What is the formula for calculating the effects of lens closeness on light refraction?

The formula for calculating the effects of lens closeness on light refraction is n1(sinθ1) = n2(sinθ2), where n1 is the refractive index of the first medium (usually air), θ1 is the angle of incidence, n2 is the refractive index of the second medium (usually the lens), and θ2 is the angle of refraction.

3. How does the curvature of a lens affect light refraction?

The curvature of a lens plays a significant role in light refraction. A convex lens, which is thicker in the middle and thinner at the edges, causes light rays to converge and focus at a point. A concave lens, which is thinner in the middle and thicker at the edges, causes light rays to diverge. This difference in curvature results in different amounts of refraction.

4. What is the relationship between lens thickness and light refraction?

The thickness of a lens also affects light refraction. A thicker lens will cause more refraction than a thinner lens, as the light has a longer distance to travel through the lens. This is why thicker lenses are used for stronger prescriptions in glasses, as they can bend light more to correct vision.

5. How can the effects of lens closeness on light refraction be applied in real-world situations?

The effects of lens closeness on light refraction have many practical applications. For example, in photography, adjusting the distance between the lens and the subject can change the focus and depth of field in an image. In microscopy, the use of oil immersion lenses allows for greater magnification by reducing the distance between the lens and the specimen. Understanding these principles can also help in the design of optical instruments, such as telescopes and microscopes.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
453
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
883
Replies
11
Views
530
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
826
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
477
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top