# Position of Final Image in Concave Mirror with Glass Block

• Volcano
In summary: Thank you very much again. Greatly apreciated.In summary, the conversation discusses the calculation of the final position of an image formed by a light source placed at the center of curvature of a concave mirror, with a glass block of thickness 60 cm and refractive index 4/3 placed in front of it. Two different results were obtained, one being at the center of curvature and the other at a distance of 105 cm from the mirror. The correct result is determined to be at the center of curvature due to the symmetry of Snell's law when the separation between the block and mirror is parallel to both surfaces. The mistake in the calculation may have been due to assuming that the focal length remains the same in the
Volcano

## Homework Statement

A glass block of thickness 60 cm and refractive index 4/3 is placed in front of a concave mirror. A light source is then placed on the center of curvature of the concave mirror. Find the position of the final image formed. (Focal length is 30 cm)

## Homework Equations

##\frac{1}{f} = \frac{1}{p} + \frac{1}{q}##

##Shift = d (1 - \frac{1}{n})##

## The Attempt at a Solution

I found two different results as final position of the image:

1. At the center of curvature. Because there is no enough space to shift for beams which coming from the source.

2. Its distance from concave mirror is 105 cm. (Calculations are below)

##Shift = 60-60.(\frac{3}{4})=15 cm##

At first the effective distance of object (light source) is shifted towards the mirror by an amount of 15 cm. Then, for concave mirror

##\frac{1}{f} = \frac{1}{p} + \frac{1}{q}##

##\frac{1}{30} = \frac{1}{60-15} + \frac{1}{q} => q= 90 cm##

But since the beams will shift (15 cm), the final position of image will be at 90 +15 = 105 cm.

Which is correct, if any :(

Last edited:
Volcano said:
At the center of curvature. Because there is no enough space to shift for beams which coming from the source.
the reason may not be space -availability;
draw ray diagrams then you can get the real picture

Volcano
Volcano said:

## Homework Statement

A glass block of thickness 60 cm and refractive index 4/3 is placed in front of a concave mirror. A light source is then placed on the center of curvature of the concave mirror. Find the position of the final image formed. (Focal length is 30 cm)

## Homework Equations

##\frac{1}{f} = \frac{1}{p} + \frac{1}{q}##

##Shift = d (1 - \frac{1}{n})##

## The Attempt at a Solution

I found two different results as final position of the image:

1. At the center of curvature. Because there is no enough space to shift for beams which coming from the source.

2. Its distance from concave mirror is 105 cm. (Calculations are below)

##Shift = 60-60.(\frac{3}{4})=15 cm##

At first the effective distance of object (light source) is shifted towards the mirror by an amount of 15 cm. Then, for concave mirror

##\frac{1}{f} = \frac{1}{p} + \frac{1}{q}##

##\frac{1}{30} = \frac{1}{60-15} + \frac{1}{q} => q= 90 cm##

But since the beams will shift (15 cm), the final position of image will be at 90 +15 = 105 cm.

Which is correct, if any :(

draw ray diagrams then write equations -that will be correct procedure-otherwise the equations can not tell you the correctness/otherwise of conclusions.

I don't think there will be an image shift, since although the beams are refracted as they enter the block, to the mirror they still seem to originate from its centre of curvature. Same happens when they exit again.

Volcano

Is it correct?

I can not draw ray diagrams for the second one (105 cm).

I think at the block mirror interface one would get refraction both ways which leads to complications.
Assume a slight separation between the block and mirror to draw the ray diagram.

Last edited:
Volcano

Slightly big separation but is it ok?

Volcano said:
Is it correct?

I can not draw ray diagrams for the second one (105 cm).

more or less correct... only the curvature of mirror should be circular and the rays are radial so that its a normal incidence making an angle 90 degrees with the element of the surface -tangential at the point of incidence.
so one does not get refraction -only the light rays will be moving slower; the surface of the medium is also spherical and rays are coming out at centre of curvature.
Volcano said:
Slightly big separation but is it ok?

i think such separaations are not in the orginal problem.
more over the shape of refracting medium is also convex of the same radius of curvature as mirror- so rays will be normally going out to mirror and can not refract at perpendicular incidence from medium to air however thin it may be.

Volcano
It seems it does not make a difference to the problem whether there is a thin separation or not. Due to the symmetry of Snell's law the beam still ends up being reflected back at the original incident angle inside of the block.

Volcano and drvrm
Is this?

More like this since the beam arrives along the normal to the surfaces

Volcano
andrevdh said:
More like this since the beam arrives along the normal to the surfaces
View attachment 98206
Ah! I got it. You meant that the separation between the block and mirror should be parallel to both surfaces. Thank you very much. Greatly apreciated.

The mistake that you made in the calculation might be due to the fact that you assumed that f stays the same in the block.
For instance the f lengthens if a lens is placed in water due to the fact that the light slows down, that is less refraction takes place in water than in air.

Volcano
andrevdh said:
The mistake that you made in the calculation might be due to the fact that you assumed that f stays the same in the block.
For instance the f lengthens if a lens is placed in water due to the fact that the light slows down, that is less refraction takes place in water than in air.
Definetly. I assumed, that stays the same place and the block as a rectangular prism.

## 1. What is a concave mirror?

A concave mirror is a curved mirror that bulges inward in the middle and reflects light in a converging manner. This means that parallel rays of light that hit the mirror will be reflected towards a central focal point.

## 2. What is a glass block?

A glass block is a solid piece of glass with flat, smooth sides. It is commonly used in construction for windows or walls, and can also be used in science experiments to demonstrate the properties of light.

## 3. How does a concave mirror affect light passing through a glass block?

A concave mirror placed behind a glass block will refract the light passing through the block. This means that the light rays will bend as they pass through the block, and then be reflected by the mirror. The amount of refraction and reflection will depend on the angle at which the light hits the block and the curvature of the mirror.

## 4. What is the purpose of using a concave mirror and glass block together in an experiment?

The concave mirror and glass block can be used together to demonstrate the principles of refraction and reflection of light. By manipulating the angle and position of the glass block and mirror, one can observe how light behaves when passing through different mediums and how it can be manipulated.

## 5. Can the concave mirror and glass block be used to create a magnifying effect?

Yes, the concave mirror and glass block can be used to create a magnifying effect. By placing an object in front of the glass block, the light from the object will pass through the block and be reflected by the mirror, creating a magnified image of the object. This is the principle behind magnifying glasses and some types of telescopes.

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