The power of a lens is +5 D. This lens is cut vertically into two equ…

The correct answer is Option (3) → \( 80 \) cm behind the lens

Cutting the lens perpendicular to its principal axis gives two plano-convex halves. Taking it as equiconvex, each half has one curved surface, so by the lens maker’s formula its focal length is doubled.

    \( \displaystyle \frac{1}{f} = (\mu-1)\left(\frac{1}{R}+\frac{1}{R}\right) = \frac{2(\mu-1)}{R} \)

    \( \displaystyle \frac{1}{f^{\prime}} = \frac{\mu-1}{R} = \frac{1}{2f} \)

    \( \displaystyle f = \frac{1}{P} = \frac{1}{5\,\mathrm{D}} = 0.20\,\mathrm{m} = 20\,\mathrm{cm} \)

    \( f^{\prime} = 2f = 40\,\mathrm{cm} \)

With \( u = -80 \) cm (Cartesian sign convention):

    \( \displaystyle \frac{1}{v}-\frac{1}{u} = \frac{1}{f^{\prime}} \)

    \( \displaystyle \frac{1}{v} = \frac{1}{40}+\frac{1}{-80} = \frac{1}{80} \)

    \( v = +80\,\mathrm{cm} \)

A positive \( v \) means a real image 80 cm behind the lens.

Shortcut: The object is at \( 80\,\mathrm{cm} = 2f^{\prime} \), so the image forms at \( 2f^{\prime} \) on the other side.

Note: The question does not say how the lens is cut. The answer given by NTA takes the cut perpendicular to the principal axis, which gives two plano-convex lenses of focal length 40 cm each. If the lens is instead cut along its principal axis, each half keeps the focal length of 20 cm and the image forms about 26.7 cm behind the lens, which is not among the options.