A screen is placed 40 cm away from an illuminated object. A converging lens is placed between the object and the screen and it is attempted to form an image of the object on the screen. If no such position of the lens can be found, the focal length of the lens
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → must be greater than 10 cm
An image on the screen is a real image. So the object and the image lie on opposite sides of the lens. Their distances from the lens add up to the object-screen distance \( D=40\,\mathrm{cm} \). A lens position exists only if the lens formula gives a real value for it.
Let the lens be at a distance \( x \) from the object, so \( u=-x \) and \( v=D-x \). The lens formula \( 1/v-1/u=1/f \) gives:
\( \displaystyle \frac{1}{D-x}+\frac{1}{x}=\frac{1}{f} \)
\( \Rightarrow\; x^2-Dx+Df=0 \)
This quadratic has a real root \( x \) only when its discriminant is zero or positive:
\( D^2-4Df\ge0 \)
\( \Rightarrow\; D\ge4f \)
\( \begin{aligned} f &\le \frac{D}{4} \\ &= \frac{40\,\mathrm{cm}}{4} \\ &= 10\,\mathrm{cm} \end{aligned} \)
So an image can be formed on the screen for any \( f\le10\,\mathrm{cm} \). For \( f=10\,\mathrm{cm} \), it forms with the lens midway, at \( u=-20\,\mathrm{cm} \) and \( v=20\,\mathrm{cm} \). Therefore, if no position of the lens works, the focal length must be greater than 10 cm.