By using a convex lens, a student obtained a sharp image of his class-room window grill on a screen. In which direction should he move the lens to focus a distant tree instead of the grill ?

According to the lens formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

1. For the window grill (closer object): The object distance $u$ is finite. The image distance $v_{\text{grill}}$ is greater than the focal length ($v_{\text{grill}} > f$). The screen is placed at this distance $v_{\text{grill}}$ to get a sharp image.
2. For the distant tree (object at infinity): The incoming rays are parallel, and the image is formed at the principal focus of the lens ($v_{\text{tree}} \approx f$).
3. Since $f < v_{\text{grill}}$, the image of the distant tree forms closer to the lens than that of the window grill.
4. Since the screen's position is fixed, to bring the image of the distant tree into sharp focus on the screen, the lens must be moved closer to the screen. This decreases the lens-to-screen distance to match the focal length ($f$).

• Why Option A is correct: Moving the lens towards the screen decreases the distance between them, aligning the screen with the focal plane where the distant tree's image is formed.
• Why Option B is incorrect: Moving the lens away from the screen increases the distance, making the image of the distant tree highly blurred.
• Why Options C and D are incorrect: These movements do not help in focusing the image on the screen.