Physics Concave lens Question Help please

Question

An object is located to the left of a concave lens whose focal length is -35 cm. The magnification produced by the lens is m1 = 0.36.

A. To decrease the magnification to  m2= 0.23, should the object be moved closer to the lens or farther away?

B. Calculate the distance through which the object should be moved.

Part A

I answered farther away, which was correct.

Part B

I did it like this and got 97.2 as my answer but it is not correct. I have 1 attempt left so please walk me through it

1/f=1/u(1/0.36-1)

1/f=1/u(25/9)

u= 25(-35)/9

u= 97.2 cm

Answer 1

The thin lens equation is 1/f = 1/d_i + 1/d_o

You also know that for case 1 you have d_i1 = -.35d_o1 and for case 2 d_i2 = -.23d_o2 (The d_i are clearly <0 and virtual.

What you want is d_o2- d_o1 (How far it needs to be moved)

For case 1 as you calculated

1/f = 1/d_o - 1/(.36d_o) = 1/d_o * (1 - 1/ .36) (notice the sign difference f<0 and d_o is >0

1/d_o = (-1/35)/(1-1/.36) = .01607 and d_o1 = 62.22 cm

Use the exact same solution for d_o1 with .23 instead of .36 in the last step.

Take the difference between the two object distances.

Tips - either stay in fractions or decimals. Be careful that you are clear what steps you use in the algebra. The 25/9 is 1/.36, not (1/.36 -1) = 16/9 .