Q.The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x.

[!TLDR] Equal optical paths require $n_1d_1=n_2d_2$; solving $1.6\times3=1.25\times x$ giv …

Two optical paths are equal (produce the same phase change) exactly when $n_1d_1=n_2d_2$ (Section 7.8.5's optical path definition, $\text{optical path}=n\times\text{actual distance}$). Given: distance $d_1=3$ cm in flint glass ($n_1=1.6$), and an equal optical path over distance $x$ cm in a second medium ($n_2=1.25$). Setting the two optical paths equal: $n_1d_1=n_2x$, i.e. $1.6\times3=1.25\times x$, so $4.8=1.25x$, giving $x=4.8/1.25=3.84$ cm -- matching the book's own printed answer exactly. Physically, since the second medium (n=1.25) is LESS dense/refractive than flint glass …

Equate the two optical paths, $n_1d_1=n_2d_2$, and solve directly …

Setting the two ACTUAL physical distances equal (or otherwise ignoring the refractive indices entirely), rather than correctly equating the two OPTICAL paths ($n\times\text{distance}$) …