Ray Optics Chapter-Wise Test 5

Object distance: \( u = -6 \, \text{cm} \) (virtual object), \( f = 12 \, \text{cm} \).

Lens formula: \( \frac{1}{v} - \frac{1}{-6} = \frac{1}{12} \Rightarrow \frac{1}{v} + \frac{1}{6} = \frac{1}{12} \).

\( \frac{1}{v} = \frac{1}{12} - \frac{1}{6} = \frac{1 - 2}{12} = \frac{-1}{12} \).

\( v = -12 \, \text{cm} \) (12 cm to the left).

Refractive index: \( n = \frac{\sin \left( \frac{A + D_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

\( A = 30^\circ \), \( D_m = 20^\circ \).

\( n = \frac{\sin \left( \frac{30 + 20}{2} \right)}{\sin \left( \frac{30}{2} \right)} = \frac{\sin 25^\circ}{\sin 15^\circ} \).

\( \sin 25^\circ \approx 0.423 \), \( \sin 15^\circ \approx 0.259 \).

\( n = \frac{0.423}{0.259} \approx 1.63 \).

A convex lens exhibits chromatic aberration with white light because different wavelengths refract by varying amounts due to the lens’s material having a wavelength-dependent refractive index. This causes colored fringes as focal points differ for each color, blurring the image.

Snell’s law: \( n_1 \sin i = n_2 \sin r \).

Glass (\( n_1 = 1.62 \)), water (\( n_2 = 1.33 \)), \( i = 50^\circ \).

\( 1.62 \times \sin 50^\circ = 1.33 \times \sin r \).

\( \sin 50^\circ \approx 0.766 \Rightarrow 1.62 \times 0.766 \approx 1.241 \Rightarrow 1.33 \sin r = 1.241 \).

\( \sin r = \frac{1.241}{1.33} \approx 0.933 \Rightarrow r = \sin^{-1}(0.933) \approx 68.9^\circ \).

Snell’s law: \( n_1 \sin i = n_2 \sin r \).

Air (\( n_1 = 1 \)), water (\( n_2 = 1.33 \)), \( i = 50^\circ \).

\( 1 \times \sin 50^\circ = 1.33 \times \sin r \).

\( \sin 50^\circ \approx 0.766 \Rightarrow 0.766 = 1.33 \sin r \Rightarrow \sin r = \frac{0.766}{1.33} \approx 0.576 \).

\( r = \sin^{-1}(0.576) \approx 35.1^\circ \).

Magnification: \( m = 1 + \frac{D}{f} \).

\( D = 25 \, \text{cm} \), \( f = 5 \, \text{cm} \).

\( m = 1 + \frac{25}{5} = 1 + 5 = 6 \).

\( f_1 = 50 \, \text{cm} \), \( f_2 = -25 \, \text{cm} \).

\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{50} + \frac{1}{-25} = \frac{1}{50} - \frac{2}{50} = \frac{1 - 2}{50} = \frac{-1}{50} \).

\( f = -50 \, \text{cm} \) (diverging system).

Focal length: \( f = -6 \, \text{cm} \), \( u = -12 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-12} = \frac{1}{-6} \Rightarrow \frac{1}{v} = \frac{1}{-6} + \frac{1}{12} = \frac{-2 + 1}{12} = \frac{-1}{12} \).

\( v = -12 \, \text{cm} \) (real image).

When a convex lens is in a medium with the same refractive index, there is no difference in refractive indices between the lens and the medium. Without this difference, light does not bend at the interfaces, and the lens acts like a flat plate, failing to converge or diverge rays to form an image.

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

\( n = 1.55 \), \( R_1 = 30 \, \text{cm} \), \( R_2 = -30 \, \text{cm} \).

\( \frac{1}{f} = (1.55 - 1) \left( \frac{1}{30} - \frac{1}{-30} \right) = 0.55 \left( \frac{1}{30} + \frac{1}{30} \right) = 0.55 \times \frac{2}{30} = \frac{1.1}{30} \).

\( f = \frac{30}{1.1} \approx 27.27 \, \text{cm} \).

Objective magnification: \( m_o = \frac{L}{f_o} = \frac{20}{1.5} \approx 13.33 \).

Eyepiece magnification: \( m_e = \frac{D}{f_e} = \frac{25}{5} = 5 \).

Total magnification: \( m = m_o \times m_e = 13.33 \times 5 \approx 66.65 \approx 67 \).

Power: \( P = \frac{1}{f} \) (in meters).

\( P = +4 \, \text{D} \Rightarrow 4 = \frac{1}{f} \Rightarrow f = \frac{1}{4} = 0.25 \, \text{m} = 25 \, \text{cm} \).

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

\( n = 1.5 \), \( R_1 = 40 \, \text{cm} \), \( R_2 = -40 \, \text{cm} \).

\( \frac{1}{f} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right) = 0.5 \left( \frac{1}{40} + \frac{1}{40} \right) = 0.5 \times \frac{2}{40} = \frac{1}{40} \).

\( f = 40 \, \text{cm} \).

When a convex lens is in a medium with a higher refractive index than the lens material, the refractive index difference reverses (medium > lens), causing the lens to diverge light instead of converging it. This occurs because the rays bend away from the normal at the interfaces.

Focal length: \( f = -5 \, \text{cm} \), \( u = -10 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-10} = \frac{1}{-5} \Rightarrow \frac{1}{v} = \frac{1}{-5} + \frac{1}{10} = \frac{-2 + 1}{10} = \frac{-1}{10} \).

\( v = -10 \, \text{cm} \).

Magnification: \( m = -\frac{v}{u} = -\frac{-10}{-10} = -1 \).

Image height: \( h' = m \times h = -1 \times 2 = -2 \, \text{cm} \) (inverted).

Magnitude = \( 2 \, \text{cm} \).