MHT CET · Physics · Ray Optics
A ray of light is incident normally on a glass slab to thickness 5 cm and refractive index 1.6. The time taken to travel by a ray from source of light to surface of slab is same as to travel through glass slab. The distance of source from the surface is
- A 5 cm
- B 8 cm
- C 12 cm
- D 24 cm
Answer & Solution
Correct Answer
(B) 8 cm
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_1=\text { time taken to reach from source}\) \(\text{to surface. } \)
\( \mathrm{s}_1=\text { distance of source from surface. } \)
\( \mathrm{T}_2=\text { time taken to travel through glass slab } \)
\( \because \mathrm{s}_2=\text { distance travelled in slab. } \)
\( \mathrm{T}_1 =\mathrm{T}_2 \)
\( \frac{\mathrm{~s}_1}{\mathrm{c}}=\frac{\mathrm{s}_2}{\mathrm{v}} \ldots(\because \text { Time }=\) \(\text { Distance/velocity }) \)
\( \therefore \frac{\mathrm{s}_1}{\mathrm{c}}=\frac{\mathrm{s}_2 \mu}{\mathrm{c}} \ldots(\because \mu=\mathrm{c} / \mathrm{v}) \)
\( \therefore \mathrm{s}_1=\mathrm{s}_2 \times \mu=5 \times 1.6=8 \mathrm{~cm}\)
\( \mathrm{s}_1=\text { distance of source from surface. } \)
\( \mathrm{T}_2=\text { time taken to travel through glass slab } \)
\( \because \mathrm{s}_2=\text { distance travelled in slab. } \)
\( \mathrm{T}_1 =\mathrm{T}_2 \)
\( \frac{\mathrm{~s}_1}{\mathrm{c}}=\frac{\mathrm{s}_2}{\mathrm{v}} \ldots(\because \text { Time }=\) \(\text { Distance/velocity }) \)
\( \therefore \frac{\mathrm{s}_1}{\mathrm{c}}=\frac{\mathrm{s}_2 \mu}{\mathrm{c}} \ldots(\because \mu=\mathrm{c} / \mathrm{v}) \)
\( \therefore \mathrm{s}_1=\mathrm{s}_2 \times \mu=5 \times 1.6=8 \mathrm{~cm}\)
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