The position of the image, if the object moved to new position 20.0 cm from the mirror is 5.7372 cm.
As per the lens equation,
[tex]\frac{1}{p1} +\frac{1}{q1} =\frac{1}{f}[/tex]
Since, p1 is the position of object initially, q1 is the position of the image initially and f is the focal length of the lens.
[tex]\frac{1}{10.4} +\frac{1}{7.8} =\frac{1}{f}[/tex]
[tex]0.0961+0.1282=\frac{1}{f}[/tex]
[tex]f=4.458[/tex] is the focal length of the given lens.
Then the object moved to second position of p2=20.0 cm from the mirror.
So, [tex]\frac{1}{p2} +\frac{1}{q2} =\frac{1}{f}[/tex]
[tex]\frac{1}{20} +\frac{1}{q2} =\frac{1}{4.458}[/tex]
[tex]\frac{1}{q2}=0.2243-\frac{1}{20}[/tex]
[tex]\frac{1}{q2}=0.1743[/tex]
Hence, q2=5.7372 cm is the new position of the image.
Initially, the object placed 10.40 cm from the mirror, then the image formed at 7.80 cm, then the object moved to 20.0 cm from the mirror, then the new image formed at 5.74 cm.
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