An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Given: Object height (h) = 5 cm, Object distance (u) = -25 cm, Focal length (f) = +10 cm
Position of image: Using lens formula: 1/v - 1/u = 1/f
1/v - 1/(-25) = 1/10
1/v + 1/25 = 1/10
1/v = 1/10 - 1/25 = (5-2)/50 = 3/50
v = 50/3 = 16.67 cm
The image is formed 16.67 cm on the opposite side of the lens.
Size of image: Magnification m = v/u = 50/3 ÷ (-25) = -2/3
Image height h' = m × h = -2/ = -3.33 cm
Nature of image: The image is real, inverted and diminished (3.33 cm tall).
Ray Diagram: Draw a convex lens with principal axis. Mark F₁ and F₂ at 10 cm from lens, and 2F₁ and 2F₂ at 20 cm. Place object of height 5 cm at 25 cm (beyond 2F₁). Draw a ray parallel to principal axis, it refracts through F₂. Draw another ray through optical centre O passing straight. The two rays meet between F₂ and 2F₂, forming a real, inverted, diminished image.
Explanation
This is a standard convex lens numerical problem. The object is placed beyond 2F₁ (25 cm > 20 cm), so according to Table 9.4, the image must form between F₂ and 2F₂, be real, inverted and diminished. The calculations confirm this: v = 16.67 cm lies between F₂ (10 cm) and 2F₂ (20 cm). The negative magnification (-2/3) confirms inverted image, and |m| < 1 confirms diminished size.
Solution Steps
Step 1: Identify given values - h = 5 cm, u = -25 cm, f = +10 cm
Step 2: Apply lens formula 1/v - 1/u = 1/f to find v
Step 3: Calculate v = 50/3 = 16.67 cm (positive means real image on opposite side)
Step 4: Calculate magnification m = v/u = -2/3
Step 5: Find image height h' = m × h = -3.33 cm (negative means inverted)
Step 6: Determine nature - real (v positive), inverted (m negative), diminished (|m| < 1)