The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against each other to form a common convex lens. The distance from this lens at which an object must be placed to obtain a real inverted image with magnification equal to unity is

Question

The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against each other to form a common convex lens. The distance from this lens at which an object must be placed to obtain a real inverted image with magnification equal to unity is (A) 80 cm (B) 40 cm (C) 20 cm (D) 160 cm

Tardigrade Admin · Accepted Answer

Focal length of the combination lens, (1/f)=(1/f1)+(1/f2)=(1/40)+(1/40) ∴ f=20 cm Given, m=1 m=(f/u-f) 1=(20/u-20) u-20=20 u=40 cm

Q. The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against each other to form a common convex lens. The distance from this lens at which an object must be placed to obtain a real inverted image with magnification equal to unity is

80 cm

40 cm

20 cm

160 cm

Focal length of the combination lens, $\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{40} + \frac{1}{40}$ $∴$ $f = 20 c m$ Given, $m = 1$ $m = \frac{f}{u - f}$ $1 = \frac{20}{u - 20}$ $u - 20 = 20$ $u = 40 c m$

Q. The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against each other to form a common convex lens. The distance from this lens at which an object must be placed to obtain a real inverted image with magnification equal to unity is

80 cm

40 cm

20 cm

160 cm

Focal length of the combination lens, f1​=f1​1​+f2​1​=401​+401​ ∴ f=20cm Given, m=1 m=u−ff​ 1=u−2020​ u−20=20 u=40cm

Focal length of the combination lens, $\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{40} + \frac{1}{40}$ $∴$ $f = 20 c m$ Given, $m = 1$ $m = \frac{f}{u - f}$ $1 = \frac{20}{u - 20}$ $u - 20 = 20$ $u = 40 c m$