(a) The position of image, and

(b) The focal length of the lens.

Solution:

According to the question,

Distance of the object u is – 30 cm

And the concave lens forms an erect image of 1/3

We know that the expression for the magnification is –

m = h’/ h

Therefore, m = 1 / 3

We know that the expression for magnification is

m = h’ / h

Also, m = v / u

Combining above two expressions, we get ==> h’ / h = v / u

Upon substituting values => 1 / 3 = v / -30

Therefore, v = -10 cm

Expression for the Lens formula is

1 / f = 1 / v – 1 / u

Upon substituting values => 1 / f = 1 / – 15

Upon solving we get f = -15 cm

Thus, the focal length is 15 cm and the image is formed at 10 cm from the lens.