Answer:
No, the image will be formed at the same position. From lens maker's formula, \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right],\] it is clear that when we interchange \[{{R}_{1}}\] and \[{{R}_{2}}\], the magnitude of \[f\] remains the same.
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