A concave 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 concave 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

m=(m/f+u) -2=((1/3)/(1)3+u) -(2/3)-2 u=(1/3) or -2 u=(1/3)+(2/3)=1 or u=-(1/2) m =-0.5 m

Q. A concave 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

$m = \frac{m}{f + u}$
$- 2 = \frac{\frac{1}{3}}{\frac{1}{3} + u}$
$- \frac{2}{3} - 2 u = \frac{1}{3}$
or $- 2 u = \frac{1}{3} + \frac{2}{3} = 1$ or $u = - \frac{1}{2} m = - 0.5 m$

Q. A concave 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

m=f+um​ −2=31​+u31​​ −32​−2u=31​ or −2u=31​+32​=1 or u=−21​m=−0.5m

Q. A concave 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

$m = \frac{m}{f + u}$
$- 2 = \frac{\frac{1}{3}}{\frac{1}{3} + u}$
$- \frac{2}{3} - 2 u = \frac{1}{3}$
or $- 2 u = \frac{1}{3} + \frac{2}{3} = 1$ or $u = - \frac{1}{2} m = - 0.5 m$