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An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

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For a given object position if the focal length of the lens does not change, the image position remains unchanged.

By lens maker's formula,

\frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )          

For this position R_{1} is positive

and R_{2} is negative. Hence focal length at this position

\frac{1}{f_{1}}=\left ( \mu -1 \right )\left ( \frac{1}{\left ( +R_{1} \right )}-\frac{1}{\left ( -R_{2} \right )} \right )=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )

Now the lens is reversed,

At this position, R_{2} is positive and R_{1} is negative. Hence focal length at this position is

\frac{1}{f_{2}}=\left ( \mu -1 \right )\left ( \frac{1}{\left ( +R_{2} \right )}-\frac{1}{\left ( -R_{1} \right )} \right )=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )

We can observe the focal length of the lens does not change in both positions, hence the image position remains unchanged.

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