Question:medium

Show the position of focus (f) in ray diagram for convex mirror. At what distance from a convex mirror of focal length 2.0 m should a boy stand so that his image has a height equal to half the original height?
OR
What is wavefront? Explain the refraction of waves with the help of Huygens' principle.

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For the convex mirror use \( m = -v/u = +\tfrac{1}{2} \) together with the mirror formula \( \tfrac{1}{v} + \tfrac{1}{u} = \tfrac{1}{f} \) and \( f = +2.0\,\text{m} \). For the OR part, a wavefront is a surface of constant phase, and Huygens' secondary-wavelet envelope gives \( \sin i/\sin r = v_1/v_2 \).
Updated On: Jul 10, 2026
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Solution and Explanation

Option 1: Convex mirror using the m = f/(f - u) form

Step 1: A convex mirror is a diverging mirror. Its principal focus F is the virtual point behind the mirror from which the reflected rays of an incident parallel beam appear to originate. Hence in a labelled ray diagram both F and the centre of curvature C are marked behind the reflecting surface, with focal length \( f = +2.0\,\text{m} \).

Step 2: For any mirror the linear magnification can be written directly in terms of focal length and object distance as \[ m = \frac{f}{f - u} \] The required image is erect and half sized, so \( m = +\dfrac{1}{2} \).

Step 3: Substitute the known values: \[ \frac{1}{2} = \frac{2.0}{2.0 - u} \]
Step 4: Cross multiply: \[ 2.0 - u = 2 \times 2.0 = 4.0 \Rightarrow u = 2.0 - 4.0 = -2.0\,\text{m} \]
Step 5: The magnitude of the object distance is 2.0 m, so the boy must stand 2.0 m from the mirror. Image position \( v = -m\,u = -(0.5)(-2.0) = +1.0\,\text{m} \), i.e. 1.0 m behind the mirror, virtual, erect and half height.
\[\boxed{\text{Boy stands }2.0\ \text{m from the mirror}}\]

Option 2: Wavefront and Huygens refraction (physical description)

Step 1: Imagine dropping a stone into still water. The moving circular crest along which every water particle is at the same stage of vibration is a wavefront. Formally, a wavefront is the continuous surface joining all particles of a medium that are in the same phase of oscillation at one instant, and energy travels at right angles to it.

Step 2: Huygens pictured each such surface as a nursery of tiny sources: (a) every point of the present wavefront emits spherical secondary wavelets moving forward at the wave speed of the medium; (b) after a time \( t \), the envelope that just touches all these wavelets on their forward side is the position of the advanced wavefront.

Step 3: Consider a plane wavefront falling obliquely on the flat interface between a rarer medium (wave speed \( v_1 \)) and a denser medium (wave speed \( v_2 \), with \( v_2 < v_1 \)). The edge of the wavefront still in medium 1 keeps moving faster than the edge that has already entered medium 2.

Step 4: In equal time \( t \), the medium-1 wavelet advances \( v_1 t \) while the medium-2 wavelet advances only \( v_2 t \). Because one edge lags behind the other, the wavefront swings round and its direction of travel turns towards the normal. Taking the two right triangles that share the interface segment AC as hypotenuse gives \( \sin i = v_1 t / AC \) and \( \sin r = v_2 t / AC \).

Step 5: Their ratio cancels \( t \) and \( AC \): \[ \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = {}_{1}n_{2} \] a constant for the given pair of media, which is exactly the experimental law of refraction. So Huygens' construction predicts both the bending of the wave and the identification of the refractive index with the ratio of speeds.
\[\boxed{\sin i / \sin r = v_1 / v_2 = \text{refractive index}}\]
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