Question

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.

Hint:

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 \( \sin i/\sin r = v_1/v_2 \).

Answer 1

Option 1: Convex mirror focus and image distance

Step 1 (Position of focus): In a convex mirror the reflecting surface bulges towards the object. A beam of rays parallel to the principal axis, after reflection, diverges outward. When produced backward, these reflected rays appear to meet at a single point F behind the mirror. This point is the principal focus. Because the focus is virtual and lies behind the mirror, the focal length of a convex mirror is taken as positive. Ray diagram (described): draw the convex mirror with pole P; mark the focus F and centre of curvature C on the principal axis behind the mirror; a ray parallel to the axis strikes the mirror and reflects so that its backward extension passes through F, so that PF = f.

Step 2 (Given data and sign convention): Focal length \( f = +2.0\,\text{m} \) (convex mirror). A convex mirror always forms a virtual, erect and diminished image, so the magnification is positive: \( m = +\tfrac{1}{2} \).

Step 3 (Magnification relation): \[ m = -\frac{v}{u} \] So \[ +\frac{1}{2} = -\frac{v}{u} \Rightarrow v = -\frac{u}{2} \]
Step 4 (Mirror formula): \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Substitute \( v = -u/2 \): \[ \frac{1}{-u/2} + \frac{1}{u} = \frac{1}{f} \] \[ -\frac{2}{u} + \frac{1}{u} = \frac{1}{f} \] \[ -\frac{1}{u} = \frac{1}{f} \Rightarrow u = -f = -2.0\,\text{m} \]
Step 5 (Result): The negative sign shows the boy stands in front of the mirror. Distance from mirror \( = 2.0\,\text{m} \). Check: \( v = -u/2 = +1.0\,\text{m} \) (image 1.0 m behind the mirror, virtual and erect); \( m = -v/u = -(1.0)/(-2.0) = +0.5 \), i.e. half the height.
\[\boxed{u = 2.0\ \text{m from the convex mirror}}\]

Option 2: Wavefront and refraction by Huygens' principle

Step 1 (Wavefront): A wavefront is the locus of all points of a medium that are vibrating in the same phase at a given instant. The wave advances perpendicular to the wavefront, and the distance moved per unit time equals the wave speed. A point source gives spherical wavefronts, while a very distant source gives plane wavefronts.

Step 2 (Huygens' principle): (i) Every point on a wavefront acts as a fresh source of secondary wavelets that spread forward in all directions with the wave speed of the medium. (ii) The new wavefront after a time interval is the forward tangential envelope (surface of tangency) of all these secondary wavelets.

Step 3 (Set-up for refraction): Let a plane wavefront AB in medium 1 (speed \( v_1 \)) reach a plane boundary XY separating it from medium 2 (speed \( v_2 \), with \( v_2 < v_1 \)). Let the incident wavefront make angle \( i \) with the surface and the refracted wavefront make angle \( r \).

Step 4 (Geometry): While the wavelet from B travels distance \( BC = v_1 t \) in medium 1 to reach the surface at C, the wavelet from A advances distance \( AD = v_2 t \) into medium 2. In the right triangles ABC and ADC (common hypotenuse AC): \[ \sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}, \qquad \sin r = \frac{AD}{AC} = \frac{v_2 t}{AC} \]
Step 5 (Snell's law): Dividing the two relations, \[ \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1} = \text{constant} \] This is Snell's law of refraction, so Huygens' principle explains the bending of waves at a boundary. Since \( v_2 < v_1 \), the ray bends towards the normal on entering the denser medium.
\[\boxed{\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \text{constant (Snell's law)}}\]