- #1

Aurelius120

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- Homework Statement
- Find the position of final image

- Relevant Equations
- Lens formula: $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$

Mirror Formula: $$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$

Mirror Magnification = ##\frac{-v}{u}##

Lens Magnification =##\frac{v}{u}##

I think the given solution is wrong.

The lens forms image at ##(+75,0)## which is ##25 cm## from pole of the convex mirror which acts as virtual object for mirror.

It is true that the reflected ray is rotated by ##2\theta## as in case of plane mirror. Rotation of Spherical Mirrors

But that doesn't necessarily mean that the image is rotated by ##2\theta##

Since the mirror is tilted the principal axis is tilted. It is a spherical mirror so the point object should be treated as the tip of an extended object and the height and position of head of extended object should give coordinates of point image.

According to this diagram:

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\implies \frac{1}{v}+\frac{2}{25\sqrt{3}}=\frac{1}{50}$$

Therefore $$v=\frac{50\sqrt{3}}{\sqrt{3}-4}$$ ## \text{ v is along tilted axis and measured from pole of mirror}##

Then from the formula of magnification:

$$\frac{I}{25/2}=\frac{-v}{25\sqrt{3}/2}\implies I=\frac{50}{4-\sqrt{3}}$$

Solving this triangle,

$$(|v|-I\tan 30)×\cos 30=50-x$$

$$y=\frac{I}{\cos 30}+(|v|-I\tan 30)×\sin 30$$

$$x=\frac{50(3-\sqrt{3}}{4-\sqrt{3}}\text{ ; } y=\frac{50\sqrt{3}}{4-\sqrt{3}}$$

These equations give the x and y coordinates of the image.

The answer does not match. There is yet another mistake in the solution, I think. I believe that this image is not the final image. The final image is formed after refraction of this image by lens.

So am I horribly misunderstanding the question? Am I terribly mistaken or is the solution wrong this time??

How to solve it correctly?

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