When an object is placed 63.0 cm from a certain converging lens, it forms a real image. When the object is moved to 53.0 cm from the lens...?

...

let X be the first image distance then the second is X+7.00

and the focal length is a constant in this problem

so

1/63.0 + 1/X = 1/53.0 + 1/(X+7.00)

1/X - 1/(X+7.00) = 1/53.0 - 1/63.0

((X+7.00) - X) / X(X+7.00) = 0.002995

7.00 / (X^2 + 7.00X) = 0.002995

X^2 + 7.00X = 7.00 / 0.002995 = 2337.3

X^2 + 7.00X - 2337.3 = 0

X = 44.97 cm

the positive root makes sense for a coerging lens

then

1/f = 1/63.0 + 1/44.97 = 0.03811

f = 26.2 cm

and a check

1/f = 1/53.0 + 1/(44.97+7.00) = 0.03811

f = 26.2 cm

When you get a good response,

please consider giving a best answer.

This is the only reward we get.

Rule

1/i + 1/o = 1/f