A biconvex lens forms a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would

Question

A biconvex lens forms a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would (A) Disappear (B) Remain as real image still (C) Be virtual and of the same size as the object (D) Suffer from aberrations

Tardigrade Admin · Accepted Answer

(1/f)=(μ-1)((1/R1)-(1/R2)) For biconvex lens R2=-R1 ∴ (1/f)=(μ-1)((2/R))

Q. A biconvex lens forms a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would

Disappear

Remain as real image still

Be virtual and of the same size as the object

Suffer from aberrations

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
For biconvex lens $R_{2} = - R_{1}$
$∴ \frac{1}{f} = (μ - 1) (\frac{2}{R})$

Q. A biconvex lens forms a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would

Disappear

Remain as real image still

Be virtual and of the same size as the object

Suffer from aberrations

f1​=(μ−1)(R1​1​−R2​1​) For biconvex lens R2​=−R1​ ∴f1​=(μ−1)(R2​)

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
For biconvex lens $R_{2} = - R_{1}$
$∴ \frac{1}{f} = (μ - 1) (\frac{2}{R})$