Option 1: Deriving Snell's law from wavefront geometry
Step 1: Huygens pictured a wavefront as a surface of points that are all vibrating in step. Each such point launches a small spherical wavelet forward at the local wave speed; a moment later the envelope touching all these wavelets is the advanced wavefront. This construction lets us track how a wavefront bends when the speed changes.
Step 2: Let parallel light in a rarer medium (speed \(v_1\)) meet a flat interface. Its wavefront edge \(B\) needs time \(t\) to reach the surface point \(C\), covering \(BC=v_1t\). During the same \(t\), the other edge at \(A\) has already crossed into the denser medium (speed \(v_2\)) and its wavelet has grown to radius \(AD=v_2t\).
Step 3: The refracted wavefront is the line \(CD\) tangent to that wavelet. The incidence angle \(i\) equals angle \(BAC\) and the refraction angle \(r\) equals angle \(DCA\) (angles measured from the surface, using the shared hypotenuse \(AC\)).
Step 4: Then \(\sin i=BC/AC=v_1t/AC\) and \(\sin r=AD/AC=v_2t/AC\). Their ratio drops the common \(t/AC\): \(\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}={}_1n_2\), a constant fixed only by the two media.
Step 5: Writing speeds as \(v=c/n\) gives \(n_1\sin i=n_2\sin r\), and the whole figure lies in one plane so ray, refracted ray and normal are coplanar. Both refraction laws follow.
\[ \boxed{\;\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=\text{constant}\;} \]
Option 2: Interference conditions and the YDSE numbers
Step 1: Coherence demands a locked phase relationship. Two independent lamps flicker randomly and cannot interfere steadily; splitting one wavefront into two slits keeps their phase tied, so they qualify as coherent.
Step 2: When coherent, equal-amplitude beams overlap, the bright and dark bands hold still instead of washing out. That fixed, time-independent pattern is what we call sustained interference.
Step 3: Right at the screen centre the two paths from the slits are identical, so the optical path difference vanishes for every colour. Zero path difference is perfect constructive interference, hence the central band is always bright.
Step 4: Convert data to SI: \(\beta=0.04\text{ cm}=4\times10^{-4}\text{ m}\), \(\lambda=6000\text{ \AA}=6\times10^{-7}\text{ m}\), \(D=1\text{ m}\). From \(\beta=\lambda D/d\), the slit gap is \(d=\lambda D/\beta=(6\times10^{-7})(1)/(4\times10^{-4})=1.5\times10^{-3}\text{ m}\), i.e. \(1.5\) mm.
Step 5: Bright fringes sit at integer multiples of \(\beta\); the third one lies at \(3\times0.04=0.12\) cm. Dark fringes sit at half-integer multiples; the second dark one lies at \(\tfrac{3}{2}\times0.04=0.06\) cm from the centre.
\[ \boxed{\;d=1.5\text{ mm},\;\;y_{3B}=0.12\text{ cm},\;\;y_{2D}=0.06\text{ cm}\;} \]