A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
The object is placed 30 cm from the lens.
Given: Focal length, f = -15 cm (concave lens) Image distance, v = -10 cm (virtual image on same side as object) Object distance, u = ?
Step 1: Using the lens formula: 1/v - 1/u = 1/f
Step 2: Rearranging: 1/u = 1/v - 1/f
Step 3: Substituting values: 1/u = 1/(-10) - 1/(-15) 1/u = -1/10 + 1/15 1/u = (-3 + 2)/30 1/u = -1/30
Step 4: Therefore: u = -30 cm
The object is placed 30 cm from the lens.
Magnification: m = v/u = -10/-30 = +1/3 = +0.33
The positive sign shows the image is virtual and erect. The image is one-third the size of the object.
Ray Diagram Description: Draw a concave lens with principal axis. Mark focus F on both sides. Place the object at 30 cm (beyond 2F). Draw a ray parallel to principal axis - it diverges appearing to come from F. Draw another ray through optical centre O passing straight. The diverging rays, when extended backward, meet at a point between F and O on the same side as the object, forming a virtual, erect, diminished image.
Explanation
This numerical is directly solved in Example 9.3 of the textbook. For a concave lens, both focal length and image distance are negative. The lens formula 1/v - 1/u = 1/f is used. The negative object distance indicates the object is on the same side as conventionally expected. The ray diagram shows the characteristic virtual, erect, diminished image formation by a concave lens.
Solution Steps
Step 1: Identify sign conventions - f = -15 cm (concave lens), v = -10 cm (virtual image)
Step 2: Apply lens formula: 1/v - 1/u = 1/f
Step 3: Rearrange to find u: 1/u = 1/v - 1/f
Step 4: Substitute and solve: 1/u = -1/10 + 1/15 = -1/30
Step 5: Calculate u = -30 cm, meaning object distance is 30 cm
Step 6: Calculate magnification m = v/u = -10/-30 = +1/3