In an experiment with a convex lens, the plot of the image distance ($$v'$$) against the object distance ($$\mu'$$) measured from the focus gives a curve $$v'\mu' = 225$$. If all the distances are measured in cm. The magnitude of the focal length of the lens is _____ cm.

We are given that when the image distance $$v'$$ and the object distance $$u'$$ are measured from the focus of a convex lens, a plot of $$v'$$ against $$u'$$ gives the curve $$v' u' = 225$$, with all distances in cm. We need to find the focal length of the lens.

This problem uses Newton's lens formula. To derive it, recall the standard thin lens equation $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$, where $$v$$ is the image distance and $$u$$ is the object distance, both measured from the optical centre of the lens.

Now, let us define $$u'$$ and $$v'$$ as the distances measured from the respective foci. If the object is at distance $$u'$$ from the first focus (on the object side), then $$u = -(f + u')$$ (taking the sign convention where the object is on the negative side). Similarly, if the image is at distance $$v'$$ from the second focus (on the image side), then $$v = f + v'$$.

Substituting into the lens equation: $$\frac{1}{f + v'} + \frac{1}{f + u'} = \frac{1}{f}$$.

Taking the LCM on the left side: $$\frac{(f + u') + (f + v')}{(f + v')(f + u')} = \frac{1}{f}$$.

Cross-multiplying: $$f(2f + u' + v') = (f + v')(f + u') = f^2 + f u' + f v' + v' u'$$.

Expanding the left side: $$2f^2 + f u' + f v' = f^2 + f u' + f v' + v' u'$$.

Cancelling the common terms $$f u'$$ and $$f v'$$ from both sides: $$2f^2 = f^2 + v' u'$$, which gives us $$v' u' = f^2$$.

This is Newton's formula for a thin lens: the product of the object and image distances measured from the respective foci equals the square of the focal length.

Given $$v' u' = 225$$, we have $$f^2 = 225$$. Taking the square root, $$f = \sqrt{225} = 15 \text{ cm}$$.

Hence, the magnitude of the focal length of the lens is $$\textbf{15}$$ cm.