A point source of light is kept at a depth of h in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is

Question

A point source of light is kept at a depth of h in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is (A)  (3/√7)h  (B)  (√7/3)h  (C)  (√3/7)h  (D)  (7/√3)h

Tardigrade Admin · Accepted Answer

( sin 90°/ sin C)=μ <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5db5be2fbdbbcf3367667c89cb4f4b1f-.png" /> sin C=(1/μ) ⇒ (R/√R2+h2)=(3/4) Squaring, 16 R2=9 R2+9 h2 7 R2=9 h2 ⇒ R=(3/√7) h

Q. A point source of light is kept at a depth of $h$ in water of refractive index $4/3$ . The radius of the circle at the surface of water through which light emits is

$\frac{3}{7} h$

$\frac{7}{3} h$

$\frac{3}{7} h$

$\frac{7}{3} h$

$\frac{s i n 9 0 ^{\circ}}{s i n C} = μ$

$sin C = \frac{1}{μ} \Rightarrow \frac{R}{R ^{2} + h ^{2}} = \frac{3}{4}$
Squaring, $16 R^{2} = 9 R^{2} + 9 h^{2}$
$7 R^{2} = 9 h^{2} \Rightarrow R = \frac{3}{7} h$

Q. A point source of light is kept at a depth of h in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is

7​3​h

37​​h

73​​h

3​7​h

sinCsin90∘​=μ sinC=μ1​⇒R2+h2​R​=43​ Squaring, 16R2=9R2+9h2 7R2=9h2⇒R=7​3​h

Q. A point source of light is kept at a depth of $h$ in water of refractive index $4/3$ . The radius of the circle at the surface of water through which light emits is

$\frac{3}{7} h$

$\frac{7}{3} h$

$\frac{3}{7} h$

$\frac{7}{3} h$

$\frac{s i n 9 0 ^{\circ}}{s i n C} = μ$

$sin C = \frac{1}{μ} \Rightarrow \frac{R}{R ^{2} + h ^{2}} = \frac{3}{4}$
Squaring, $16 R^{2} = 9 R^{2} + 9 h^{2}$
$7 R^{2} = 9 h^{2} \Rightarrow R = \frac{3}{7} h$