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Answer 1

Aɴsᴡᴇʀ :

• $\sf{Refractive\: index \:of\: }\bf{\purple{z}} \:\sf{w.r.t}\: \bf{\purple{x}} =$$\bf{\red{\dfrac{9}{8}}}$

Gɪᴠᴇɴ :

• $\sf{Refractive \:index\: of \:}\bf{\purple{x}}\:\sf{ w.r.t}\: \bf{\purple{y}}\sf{= \dfrac{2}{3}}$

• $\sf{Refractive \:index\: of\: }\bf{\purple{y}}\:\sf{ w.r.t}\: \bf{\purple{z}}= \sf{\dfrac{4}{3}}$

Tᴏ Fɪɴᴅ :

• $\sf{Refractive\: index \:of\: }\bf{\purple{z}}\:\sf{ w.r.t}\: \bf{\purple{x}}.$

Fᴏʀᴍᴜʟᴀ Usᴇᴅ :

• $\sf{Refractive \:index\: of\: }\bf{\purple{1}} \:\sf{w.r.t }\:\bf{\purple{2}} = \sf{n_{12} = \dfrac{n_1}{n_2}}$

Sᴏʟᴜᴛɪᴏɴ :

• $\sf{n_{xy} = \dfrac{n_x}{n_y} = \dfrac{2}{3}}$

• $\sf{n_{yz} = \dfrac{n_y}{n_z} = \dfrac{4}{3}}$

• $\sf{n_{zx} = \dfrac{n_z}{n_x}}$$\sf{= \dfrac{n_z}{n_y} \times \dfrac{n_y}{n_x}}$

$\sf{\:\:\:\:\:\:\:\:\:\:= \dfrac{3}{4} \times \dfrac{3}{2}=}$$\sf{\red{\dfrac{9}{8}}}$