A convex lens of focal length (1/3) m forms a real inverted image twice in size of the object. The distance of the object from the lens is

Question

A convex lens of focal length (1/3) m forms a real inverted image twice in size of the object. The distance of the object from the lens is (A) 0.5 m (B) 0.166 m (C) 0.33 m (D) 1 m

Tardigrade Admin · Accepted Answer

For lens, m=-2, f=+(1/3) m ⇒ (v/u)=-2 or v=-2 u As (1/v)-(1/u)=(1/f) ⇒ (1/-2 u)-(1/u)=3 (-3/2 u)=3 ⇒ u=(-1/2) m =-0.5 m

Q. A convex lens of focal length $\frac{1}{3} m$ forms a real inverted image twice in size of the object. The distance of the object from the lens is

0.5 m

0.166 m

0.33 m

1 m

For lens,
$m = - 2, f = + \frac{1}{3} m \Rightarrow \frac{v}{u} = - 2$ or $v = - 2 u$
As $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{- 2 u} - \frac{1}{u} = 3$
$\frac{- 3}{2 u} = 3 \Rightarrow u = \frac{- 1}{2} m = - 0.5 m$

Q. A convex lens of focal length 31​m forms a real inverted image twice in size of the object. The distance of the object from the lens is

0.5 m

0.166 m

0.33 m

1 m

For lens, m=−2,f=+31​m⇒uv​=−2 or v=−2u As v1​−u1​=f1​⇒−2u1​−u1​=3 2u−3​=3⇒u=2−1​m=−0.5m

Q. A convex lens of focal length $\frac{1}{3} m$ forms a real inverted image twice in size of the object. The distance of the object from the lens is

For lens,
$m = - 2, f = + \frac{1}{3} m \Rightarrow \frac{v}{u} = - 2$ or $v = - 2 u$
As $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{- 2 u} - \frac{1}{u} = 3$
$\frac{- 3}{2 u} = 3 \Rightarrow u = \frac{- 1}{2} m = - 0.5 m$