Physics Asked on March 11, 2021
Pin is kept in front of a concave mirror at a distance $x$ . Observer shifts his gaze towards the left,the inverted image of pin appears to the right. Then $f<x<2f$ .
Here I understand that the rightward shift is because of parallax. I am confused as to why it will be seen only when $f<x<2f$ , as from what I’ve seen in practical observations parallax can be observed from any position (correct me if I’m wrong).
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Parallax is observed at most positions. I say most because if you keep the object pin at the center of curvature $(x=2f)$, you will see the image moving with the object pin (no parallax). Also, if you place the object at the focus $(x=f)$, you won't get a clear image. For all other positions of the object pin, you should observe parallax.
Observer shifts his gaze towards the left,the inverted image of pin appears to the right. Then $f<x<2f$.
This statement has 2 pieces of information:
This is enough to deduce the range of $x$; because:
Correct answer by user220805 on March 11, 2021
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