An object 'O' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm s$$^{-1}$$. The position of the image from the mirror after 10 s will be at _____ cm.

We need to determine the position of the image formed by a concave mirror after an object 'O' has been moving towards it for 10 seconds.

Step 1: Identify the Initial Mirror Parameters

According to the standard Cartesian sign convention for a concave mirror:

• Radius of curvature ($$R$$) = $$-200\text{ cm}$$
• Focal length ($$f$$) = $$\frac{R}{2} = \frac{-200}{2} = -100\text{ cm}$$
• Initial object distance ($$u_{\text{initial}}$$) = $$-100\text{ cm}$$

Note: Initially, the object is placed exactly at the focal point of the mirror ($$u = f$$).

Step 2: Calculate the New Object Position after 10 Seconds

The object moves towards the mirror at a constant speed of $$2\text{ cm s}^{-1}$$.

• Distance traveled by the object in $10\text{ s}$ ($\Delta s$) = $$\text{speed} \times \text{time} = 2\text{ cm s}^{-1} \times 10\text{ s} = 20\text{ cm}$$

Since the object moves closer to the mirror, its new distance from the pole decreases in magnitude:

$$\text{New distance} = 100\text{ cm} - 20\text{ cm} = 80\text{ cm}$$

Applying the sign convention, the new object distance ($$u$$) is:

$$u = -80\text{ cm}$$

Step 3: Calculate the Image Position ($$v$$) Using the Mirror Formula

The relationship between the focal length, object distance, and image distance is given by the mirror formula:

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

Rearranging the formula to isolate $$\frac{1}{v}$$:

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

Substituting the values ($$f = -100\text{ cm}$$ and $$u = -80\text{ cm}$$):

$$\frac{1}{v} = \frac{1}{-100} - \frac{1}{-80}$$

$$\frac{1}{v} = -\frac{1}{100} + \frac{1}{80}$$

Finding a common denominator (400) to simplify the fractions:

$$\frac{1}{v} = \frac{-4 + 5}{400} = \frac{1}{400}$$

$$v = 400\text{ cm}$$

Conclusion

The positive sign indicates that a virtual image is formed behind the reflecting surface of the mirror.

Therefore, the position of the image from the mirror after 10 seconds will be at 400 cm.