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(A,C,C) $1.$ Since the wavefronts are parallel straight lines,the light rays (which are perpendicular to the wavefronts) are parallel. Thus,light trav

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(A,C,C) $1.$ Since the wavefronts are parallel straight lines,the light rays (which are perpendicular to the wavefronts) are parallel. Thus,light travels as a parallel beam in each medium. Correct option is $(A)$.$2.$ For a plane wavefront,the phase difference between two points depends on the path difference. Since the wavefronts are parallel,the distance between any two points on the same wavefront is constant. The phase difference between two points on the same wavefront is zero. Thus,$(\phi_{d}-\phi_{c}) = 0$ and $(\phi_{f}-\phi_{e}) = 0$. Therefore,$(\phi_{d}-\phi_{c}) = (\phi_{f}-\phi_{e})$. However,the question asks for the correct statement. Given $\phi_{c} 
eq \phi_{f}$,the phase difference between points on different wavefronts is non-zero. The correct relation is that the phase difference between points on the same wavefront is zero. Looking at the options,$(C)$ is the only one that holds true as both sides are zero. Correct option is $(C)$.$3.$ The distance between successive wavefronts represents the wavelength $\lambda$. From the figure,the distance between $ab$ and $cd$ (wavelength in medium-$1$,$\lambda_1$) is smaller than the distance between $ef$ and $gh$ (wavelength in medium-$2$,$\lambda_2$). Since $v = f\lambda$ and frequency $f$ remains constant during refraction,$v \propto \lambda$. Thus,$v_2 > v_1$. The speed is larger in medium-$2$ than in medium-$1$. Correct option is $(C)$.

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