Why Does Moving the Microscope Objective Create a Second Image?

[Numen]

Homework Statement

I have a transparent plate (call the thickness t and refractive index n), it has a scratch at a point A. A microscope objective (M, positive) images the point A in A' at a distance D from the objective. If we move M a certain distance d down (without changing focus, meaning D is unchanged), there will be another sharp image of the scratch.

How come?

Homework Equations

Not needed?

The Attempt at a Solution

I was thinking that it might be some reflection on the objective that reflects the light back to the plate and back to the objective at the same point as before it was moved, and then refracted towards A'. I'm not sure if I understand the problem correctly... I think it is suggesting a second image somewhere else?

 

[Numen]

Ok, I tried illustrating the problem. Please check the attachment (hope it helps).

 

[m2003]

I can provide a response to this content by explaining the phenomenon of reflection and refraction in optics. When light passes through a transparent material, such as the transparent plate in this problem, it undergoes both reflection and refraction. Reflection occurs when light bounces off the surface of the material, while refraction occurs when light passes through the material and changes direction due to a change in its speed.

In this case, the scratch on the plate acts as a point of discontinuity for the light passing through it. As the light reaches this point, some of it will be reflected back towards the objective, while the rest will continue to pass through and be refracted towards the point A'. This is why the objective is able to create a sharp image of the scratch at point A'.

When the objective is moved down a certain distance, the reflected light still follows the same path and reaches the objective at the same point as before. However, the refracted light now travels a longer distance before reaching the point A', resulting in a second sharp image of the scratch. This is because the distance between the objective and the point A' has remained the same, but the angle of refraction has changed due to the movement of the objective.

In summary, the presence of the scratch on the transparent plate causes both reflection and refraction of the light passing through it, resulting in two sharp images of the scratch when the objective is moved a certain distance down. This phenomenon can be explained using the laws of reflection and refraction in optics.

 

[kyle8921]

Your reasoning is correct. When the objective is moved down, the light from the scratch at point A will still be reflected and refracted by the plate, but now it will also be reflected by the objective itself. This reflected light will then be refracted again by the plate and will end up at the same point A' as before. This is known as a virtual image and it appears as if the scratch is located at a point behind the plate. This phenomenon is a result of the multiple reflections and refractions that occur within the system.