Verified Solution • 2025–26 Session moderate

QJEEM24Ph85f3: JEE Main 2024 Physics - Class 12 JEEM24Ph85f3 Solution: A light ray is incident on a glass slab of thickness and…

A light ray is incident on a glass slab of thickness 43 cm4 \sqrt{3} \mathrm{~cm} and refractive index 2\sqrt{2} The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ______ cm\mathrm{cm} .

(Given sin⁡15∘=0.25\sin 15^{\circ}=0.25 )

Step-by-Step Verified Solution

amp;μ=2amp;sin⁡θC=12amp;QC=45∘amp;i=QC=45∘\begin{aligned}& \mu=\sqrt{2} \\& \sin \theta_C=\frac{1}{\sqrt{2}} \\& Q_C=45^{\circ} \\& i=Q_C=45^{\circ}\end{aligned}

amp; ( ϕ ) lateral displacement =tsin⁡(i−r)cos⁡ramp;sin⁡45∘=2sin⁡ramp;⇒r=30∘amp;∴d=43sin⁡(45∘−30∘)cos⁡30∘amp;=43×1432=2\begin{aligned}& \text { ( } \phi \text { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \\& \sin 45^{\circ}=\sqrt{2} \sin r \\& \Rightarrow \quad r=30^{\circ} \\& \therefore \quad d=\frac{4 \sqrt{3} \sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\& =\frac{4 \sqrt{3} \times \frac{1}{4}}{\frac{\sqrt{3}}{2}}=2\end{aligned}

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