physics class 12

Question

the distance of the image formed by an equi-convex lens in air from it's second focus X₂= 30cm while the distance of the object from the first focus is X₁= 10cm. Calculate the focal length of the lens.

Please answer ASAP. I need it urgently. Please give the correct answer not copy pasted google answer. Please....

Answer 1

\(\frac 1v - \frac 1u = \frac 1f\)

\(\frac 1{30+ f} -\frac 1{-(10 + f)} = \frac 1f\)

\(\frac 1{30+ f} +\frac 1{10 + f} = \frac 1f\)

\(\frac {(40 + 2f)}{(30 + f)(f + 10)} = \frac 1f\)

\(40f + 2f^2 = 40f + f^2 + 300\)

\(f^2 = 300\)

\(f = 17.3\) cm

Comments

could u tell me how u got (40+2f) in the 3rd step??

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1/(30+f) + 1/(10+f) = 1/f

10+f+30+f / (30+f)(10+f) = 1/f

40+2f / (30+f)(10+f) = 1/f