The shape of graph between 1/u and 1/v in case of a convex lens is:

The correct answer is Option (1)

The lens formula, with the Cartesian sign convention (distances measured from the optical centre, positive in the direction of the incident light), is

    \( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; \frac{1}{v} = \frac{1}{u} + \frac{1}{f} \)

With \( 1/u \) on the x-axis and \( 1/v \) on the y-axis this is a straight line of slope +1. For a convex lens \( f \) is positive.

    \( \text{Putting } 1/u = 0\text{: } 1/v = 1/f \ (\text{positive, on the y-axis}) \)
    \( \text{Putting } 1/v = 0\text{: } 1/u = -1/f \ (\text{negative, on the x-axis}) \)

For a real object \( u \) is negative, so the graph runs from a point on the negative x-axis up to a point on the positive y-axis, at equal distances \( 1/f \) from the origin. Graph (1) shows this line, from A on the negative x-axis to B on the positive y-axis.

NCERT (2026-27):
Refraction by a Lens – “… Applying the sign convention, BO = –u, DI = +v, we get

    \( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \qquad (9.23) \)

Equation (9.23) is the familiar thin lens formula.”