Question

A small bulb is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth \( H \). The radius of the circular area of the surface of liquid, through which the light from the bulb can emerge out, is \( R \). Then \( \left( \frac{R}{H} \right) \) is:

• A. \( \frac{1}{\sqrt{n^2 - 1}} \)
• B. \( \sqrt{n^2 - 1} \)
• C. \( \frac{1}{\sqrt{n^2 + 1}} \)
• D. \( \sqrt{n^2 + 1} \)
• A. 1.3
• B. 1.4
• C. 1.5
• D. 1.6
• A. decrease by 4\%
• B. increase by 4\%
• C. decrease by 19\%
• D. increase by 19\%
• A. always real
• B. always virtual
• C. first real and then virtual
• D. first virtual and then real
• A. 10 cm
• B. 15 cm
• C. 20 cm
• D. 40 cm

Answer 1

The Correct Option is A

Step 1: Understand the setup.
The bulb is at depth \( H \) in a liquid (refractive index \( n \)). Light emerges into air (refractive index 1) within a circular area of radius \( R \), determined by the critical angle. Step 2: Apply the critical angle concept.
At the critical angle \( \theta_c \), light just emerges (angle in air = 90°). Using Snell’s law: \[ n \sin \theta_c = 1 \cdot \sin 90^\circ \quad \Rightarrow \quad \sin \theta_c = \frac{1}{n}. \] Step 3: Relate \( R \) and \( H \).
In the right triangle formed by the light ray: \[ \tan \theta_c = \frac{R}{H}. \] \[ \cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \left(\frac{1}{n}\right)^2} = \sqrt{\frac{n^2 - 1}{n^2}}, \] \[ \tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} = \frac{\frac{1}{n}}{\sqrt{\frac{n^2 - 1}{n^2}}} = \frac{1}{\sqrt{n^2 - 1}}. \] Thus: \[ \frac{R}{H} = \tan \theta_c = \frac{1}{\sqrt{n^2 - 1}}. \] Step 4: Match with options.
The expression matches option (A).

Answer 2

Step 1: Define the given quantities.
Refracting angle \( A = 60^\circ \), angle of minimum deviation \( \delta_m = 30^\circ \). Step 2: Use the formula for minimum deviation.
The refractive index \( n \) of the prism is given by: \[ n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}. \] Step 3: Substitute the values.
\[ \frac{A + \delta_m}{2} = \frac{60^\circ + 30^\circ}{2} = 45^\circ, \quad \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ. \] \[ n = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \approx 1.414. \] Step 4: Match with options.
The value \( 1.414 \) is closest to 1.4, so the answer is option (B).

Answer 3

Step 1: Use the formula for a thin prism.
For a thin prism, the angle of minimum deviation is: \[ \delta_m = (n - 1)A, \] where \( n \) is the refractive index, and \( A \) is the refracting angle. Step 2: Calculate for crown glass.
For crown glass (\( n_1 = 1.52 \)): \[ \delta_{m1} = (1.52 - 1)A = 0.52A. \] Step 3: Calculate for dense flint glass.
For dense flint glass (\( n_2 = 1.62 \)): \[ \delta_{m2} = (1.62 - 1)A = 0.62A. \] Step 4: Find the percentage change.
Change in angle: \[ \delta_{m2} - \delta_{m1} = 0.62A - 0.52A = 0.10A. \] Percentage change: \[ \frac{\delta_{m2} - \delta_{m1}}{\delta_{m1}} \times 100 = \frac{0.10A}{0.52A} \times 100 = \frac{0.10}{0.52} \times 100 \approx 19.23\%. \] Since \( \delta_{m2}>\delta_{m1} \), the angle increases by approximately 19\%. Step 5: Match with options.
The result matches option (D).

Answer 4

Step 1: Set up the refraction formula.
Light travels from air (\( n_1 = 1 \)) to glass (\( n_2 = 1.5 \)). The surface is convex toward air, so \( R \) is positive. The refraction formula is: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}. \] \[ \frac{1.5}{v} - \frac{1}{u} = \frac{1.5 - 1}{R} = \frac{0.5}{R}. \] Step 2: Initial position (\( u = -4R \)).
\[ \frac{1.5}{v} - \frac{1}{-4R} = \frac{0.5}{R}, \] \[ \frac{1.5}{v} = \frac{0.5}{R} - \frac{1}{4R} = \frac{1}{4R}, \] \[ v = 6R. \] The image is real (\( v>0 \)). Step 3: Transition point (\( u = -2R \)).
\[ \frac{1.5}{v} - \frac{1}{-2R} = \frac{0.5}{R}, \] \[ \frac{1.5}{v} = 0 \quad \Rightarrow \quad v \to \infty. \] Step 4: Closer position (\( u = -R \)).
\[ \frac{1.5}{v} - \frac{1}{-R} = \frac{0.5}{R}, \] \[ \frac{1.5}{v} = -\frac{0.5}{R}, \] \[ v = -1.5R. \] The image is virtual (\( v<0 \)). Step 5: Conclusion.
Initially (\( u = -4R \)), the image is real. As the object moves closer (\( |u|<2R \)), the image becomes virtual. Thus, the answer is option (C).

Answer 5

Step 1: Use the lensmaker’s formula.
For a lens in air: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right), \] where \( f = 10 \, \text{cm} \), \( n = 1.5 \). Step 2: Assign sign convention.
For a double-convex lens, \( R_1 = +R \) (first surface convex to the left), \( R_2 = -R \) (second surface convex to the right): \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (n - 1) \frac{2}{R}. \] Step 3: Substitute the values.
\[ \frac{1}{10} = (1.5 - 1) \frac{2}{R} = 0.5 \times \frac{2}{R}, \] \[ \frac{1}{R} = \frac{1}{10} \quad \Rightarrow \quad R = 10 \, \text{cm}. \] Step 4: Match with options.
The radius of curvature is 10 cm, which matches option (A).