Understanding Critical Angle of Incidence
The question asks for the condition that defines the critical angle of incidence, denoted as $i_c$, when light travels from a medium where it is denser to a medium where it is rarer.
Light Bending from Denser to Rarer Medium
When a ray of light travels from a denser medium (higher refractive index, $n_1$) to a rarer medium (lower refractive index, $n_2$), it bends away from the normal. The angle of refraction ($\theta_r$) is greater than the angle of incidence ($i$).
Defining the Critical Angle
As the angle of incidence ($i$) increases, the angle of refraction ($\theta_r$) also increases. The critical angle of incidence ($i_c$) is a specific angle of incidence. It is defined as the angle of incidence for which the corresponding angle of refraction is exactly $90^\circ$. At this point, the refracted ray travels along the boundary surface between the two media.
- Angle of Incidence = $i_c$
- Angle of Refraction = $\theta_r = 90^\circ$
Applying Snell's Law
Snell's Law describes the relationship between the angles and refractive indices:
$ n_1 \sin(i) = n_2 \sin(\theta_r) $
At the critical angle ($i = i_c$), the angle of refraction is $90^\circ$ ($\theta_r = 90^\circ$). Substituting these values into Snell's Law:
$ n_1 \sin(i_c) = n_2 \sin(90^\circ) $
Since $\sin(90^\circ) = 1$, the equation becomes:
$ n_1 \sin(i_c) = n_2 $
This gives the formula for the critical angle:
$ \sin(i_c) = \frac{n_2}{n_1} $
This formula is valid only when $n_1 > n_2$ (denser to rarer medium), ensuring that $\frac{n_2}{n_1} < 1$, which is necessary for $\sin(i_c)$ to have a real value.
Analyzing the Options
- Option 1: the angle of reflection is $90^\circ$ - The angle of reflection is equal to the angle of incidence. This condition doesn't define the critical angle for refraction.
- Option 2: the angle of refraction is $90^\circ$ - This is the correct definition of the critical angle of incidence ($i_c$).
- Option 3: the angle of refraction is $0^\circ$ - This occurs when the angle of incidence is $0^\circ$ (the ray is perpendicular to the surface).
- Option 4: the angle of reflection is $0^\circ$ - This happens when the angle of incidence is $0^\circ$.
Therefore, the critical angle of incidence is the angle at which the angle of refraction becomes $90^\circ$.
Important Questions from Ray Optics and Optical Instruments
A Convex mirror produces the magnification 1/3 and 1/4 when the object is placed at the points P and Q in front of the mirror.
Which of the following statements are correct?
- A. The saturation current is constant with collector plate potential for different frequencies of incident radiation.
- B. The saturation current is different with collector plate potential for different frequencies of incident radiation.
- C. The saturation current is different with collector plate potential for different intensity of incident radiation.
- D. The saturation current is constant with collector plate potential for different intensity of incident radiation.
- E. Below threshold frequency, no photoelectrons are emitted.
Choose the correct answer from the options given below:
For insulators and semiconductors, the resistance decreases with an increase in temperature because:
A ray of light passes through four transparent media with refractive index μ1, μ2, μ3, and μ4 as shown in the figure. The surfaces of all media are parallel. If BC and DE are parallel, we must have:
Light of uniform intensity shines perpendicularly on a totally absorbing surface, fully illuminating the surface. If the area of the surface is decreased, what is the effect on radiation pressure?
