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Q10:
Rays from Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of its image is equal to the size of the object?

Solution :

Step 1: Determining the Focal Length ($f$)

According to the principles of optics, light rays emanating from a distant source like the Sun are considered to be parallel to the principal axis. When these rays converge at a point after reflection from a concave mirror, that point is defined as the principal focus ($F$).

Given that the rays converge at a point $15\text{ cm}$ in front of the concave mirror:

  • Focal length ($f$) = $-15\text{ cm}$ [Per the sign convention for concave mirrors, the focus lies in front of the mirror, hence it is negative].

Step 2: Identifying the Required Object Position

The problem stipulates that the size of the image ($h_i$) must be equal to the size of the object ($h_o$), which implies a linear magnification ($m$) of $-1$.

$m = \frac{h_i}{h_o} = -1$

For a concave mirror, a real and inverted image of the same size as the object is formed if and only if the object is placed at the Center of Curvature ($C$). [Per the properties of concave mirrors: when $u = 2f$, $v = 2f$ and $m = -1$].

Step 3: Calculation of the Position ($u$)

The radius of curvature ($R$) is related to the focal length ($f$) by the formula $R = 2|f|$.

$R = 2 \times 15\text{ cm} = 30\text{ cm}$

Since the object must be placed at the center of curvature, the distance $u$ is:

$u = -30\text{ cm}$ [By sign convention, the object is placed in front of the mirror].

Step 4: Geometric Representation of the Optical System

Mirror F (-15cm) C (-30cm) Object at C

Step 5: Theoretical Verification via Mirror Formula

Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{-15} = \frac{1}{v} + \frac{1}{-30}$

$\frac{1}{v} = \frac{1}{30} - \frac{1}{15} = \frac{1 - 2}{30} = -\frac{1}{30}$

$v = -30\text{ cm}$

Magnification $m = -\frac{v}{u} = -\frac{-30}{-30} = -1$. The calculation confirms that the image size equals the object size at this position.

Final Solution: The object should be placed at 30 cm in front of the concave mirror (at the center of curvature).


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