Question

An object of height 3.6 cm is placed normally on the principal axis of a concave mirror of radius of curvature 30 cm. If the object is at a distance of 10 cm from the principal focus of the mirror, then the height of the real image formed due to the mirror is

• A. 5.4 cm
• B. 3.6 cm
• C. 1.8 cm
• D. 2.7 cm

Hint:

Always draw a rough ray diagram for mirror/lens problems. For a concave mirror, an object placed between C (-30 cm) and F (-15 cm), like at u=-25 cm, will form a real, inverted, and magnified image beyond C. This helps in verifying the signs and relative magnitude of your answer.

Answer 1

The Correct Option is A

Step 1: Newton's Formula for Spherical Mirrors:
When distances are measured from the focus, Newton's formula states:
x1 x2 = f2
where x1 is the object distance from focus, x2 is the image distance from focus, and f is the focal length.
Also, magnification m = -f / x1 or m = -x2 / f.
Actually, a simpler relation for magnification magnitude is |m| = f / x, where x is the object distance from focus.
Magnification is also given by m = hi / ho.

Step 2: Given Data:
Radius of curvature R = 30 cm.
Focal length f = R / 2 = 15 cm.
Object distance from principal focus x1 = 10 cm.
Since it is a real image, the object must be outside the focus.
So, the object is placed at 15 + 10 = 25 cm from the pole.
Using the relation for magnification magnitude:
|m| = f / x = 15 / 10 = 1.5

Step 3: Calculate Image Height:
|m| = Height of Image / Height of Object
1.5 = hi / 3.6
hi = 1.5 × 3.6 = 5.4 cm.