Ray Optics Chapter-Wise Test 15

Optical fibers use total internal reflection to transmit light. The core has a higher refractive index than the cladding, ensuring that light rays striking the boundary at angles greater than the critical angle are fully reflected, preventing loss of light intensity over distance.

Object distance: \( u = -6 \, \text{cm} \) (virtual object), \( f = -15 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{-6} = \frac{1}{-15} \Rightarrow \frac{1}{v} + \frac{1}{6} = \frac{1}{-15} \).

\( \frac{1}{v} = \frac{1}{-15} - \frac{1}{6} = \frac{-2 - 5}{30} = \frac{-7}{30} \).

\( v = -\frac{30}{7} \approx -4.29 \, \text{cm} \) (4.29 cm to the left).

Focal length: \( f = -8 \, \text{cm} \), \( u = -16 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-16} = \frac{1}{-8} \Rightarrow \frac{1}{v} = \frac{1}{-8} + \frac{1}{16} = \frac{-2 + 1}{16} = \frac{-1}{16} \).

\( v = -16 \, \text{cm} \).

Magnification: \( m = -\frac{v}{u} = -\frac{-16}{-16} = -1 \).

Image height: \( h' = m \times h = -1 \times 4 = -4 \, \text{cm} \) (inverted).

Magnitude = \( 4 \, \text{cm} \).

In a refracting telescope, the objective lens forms a real, inverted image of the distant object at its focal plane. The eyepiece magnifies this inverted image without reinverting it, so the final image remains inverted unless an additional lens or prism system is used.

Shift = \( t \left( 1 - \frac{1}{n} \right) \).

\( t = 9.6 \, \text{cm} \), \( n = 1.6 \).

Shift = \( 9.6 \left( 1 - \frac{1}{1.6} \right) = 9.6 \left( 1 - 0.625 \right) = 9.6 \times 0.375 = 3.6 \, \text{cm} \).

Lens maker’s formula: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).

\( n = 1.55 \), \( R_1 = 25 \, \text{cm} \), \( R_2 = -25 \, \text{cm} \).

\( \frac{1}{f} = (1.55 - 1) \left( \frac{1}{25} - \frac{1}{-25} \right) = 0.55 \left( \frac{1}{25} + \frac{1}{25} \right) = 0.55 \times \frac{2}{25} = \frac{1.1}{25} \).

\( f = \frac{25}{1.1} \approx 22.73 \, \text{cm} \).

Total internal reflection occurs when light travels from a denser to a rarer medium and the angle of incidence exceeds the critical angle. The critical angle depends on the refractive indices of the two media, making the angle of incidence the decisive factor for the phenomenon.

Focal length: \( f = -15 \, \text{cm} \), \( u = -30 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-30} = \frac{1}{-15} \Rightarrow \frac{1}{v} = \frac{1}{-15} + \frac{1}{30} = \frac{-2 + 1}{30} = \frac{-1}{30} \).

\( v = -30 \, \text{cm} \).

Magnification: \( m = -\frac{v}{u} = -\frac{-30}{-30} = -1 \).

Image height: \( h' = m \times h = -1 \times 5 = -5 \, \text{cm} \) (inverted).

Magnitude = \( 5 \, \text{cm} \).

Object distance: \( u = -5 \, \text{cm} \) (virtual object), \( f = -20 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{-5} = \frac{1}{-20} \Rightarrow \frac{1}{v} + \frac{1}{5} = \frac{1}{-20} \).

\( \frac{1}{v} = \frac{1}{-20} - \frac{1}{5} = \frac{-1 - 4}{20} = \frac{-5}{20} = \frac{-1}{4} \).

\( v = -4 \, \text{cm} \) (4 cm to the left).

A convex mirror reflects light such that the rays diverge after reflection. These diverging rays appear to originate from a point behind the mirror, forming a virtual image. Since the rays do not actually converge, a real image (which requires convergence) cannot be formed.