Position of Final Image in Concave Mirror with Glass Block

[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 :(

 

[drvrm]

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

 

[drvrm]

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.

 

[andrevdh]

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).

 

[andrevdh]

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.

 

[Volcano]

Slightly big separation but is it ok?

 

[drvrm]

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.

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.

 

[andrevdh]

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]

Is this?

(Sorry, I'm taking your time)

 

[andrevdh]

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

 

[Volcano]

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.

 

[andrevdh]

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]

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.