Question
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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Answer
Key concept: Thin lens formula: $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ For a given object position if focal length of the lens deos not change, the image posrition remains unchanged. 
By lens maker's formula, $\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ For this position R1 is positive and R2 is negative. Hence focal lenght at this position $\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{(+\text{R}_1)}-\frac{1}{(-\text{R}_2)}\Big)=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ 
By lens maker's formula, $\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ For this position R1 is positive and R2 is negative. Hence focal lenght at this position $\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{(+\text{R}_1)}-\frac{1}{(-\text{R}_2)}\Big)=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ Now the lens is reversed,
At this position, R2 is positive and R1 is negative. Hence focal length at this position is $\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{(+\text{R}_2)}-\frac{1}{(-\text{R}_1)}\Big)=(\mu-1)\Big(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\Big)$ We can observe the focal length of the lens does not change in both positions, hence the image position remains unchanged.Need a full question paper?
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