A ray of light makes a glancing angle of incidence of 60° with a flat glass surface. Given that the critical angle for glass is 42°, determine the angle of refraction r.

Part 1: Angle of refraction

Speed of light in air: $c = 3 \times 10^{8}$ m/s

Speed of light in glass: $v_g = 2 \times 10^{8}$ m/s

Step 1: Refractive index of glass
$n_g = \frac{c}{v_g}$
Substitute values:
$n_g = \frac{3 \times 10^{8}}{2 \times 10^{8}} = 1.5$

Step 2: Snell's law
$n_{air} \sin i = n_g \sin r$
Substitute $n_{air} = 1$, $i = 45^\circ$:
$1 \cdot \sin 45^\circ = 1.5 \sin r$
$\sin r = \frac{\sin 45^\circ}{1.5}$

Step 3: Compute $\sin 45^\circ = \frac{\sqrt{2}}{2}$
$\sin r = \frac{\frac{\sqrt{2}}{2}}{1.5} = \frac{\sqrt{2}}{2} \times \frac{2}{3} = \frac{\sqrt{2}}{3}$
$\sin r = \frac{1.414}{3} = 0.4713$

Step 4: Angle of refraction
$r = \sin^{-1}(0.4713) \approx 28^\circ$
Final answer: $r = 28^\circ$

Part 2: Magnifying power of microscope

Focal length of objective: $f_o = 0.50$ cm
Focal length of eyepiece: $f_e = 2.5$ cm
Tube length: $L = 16.0$ cm
Least distance of distinct vision: $D = 25$ cm

Step 1: Image distance for objective (normal adjustment)
$L = v_o + f_e$
Substitute values:
$16.0 = v_o + 2.5$
$v_o = 16.0 - 2.5 = 13.5$ cm

Step 2: Object distance for objective (lens formula)
$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$
$\frac{1}{u_o} = \frac{1}{f_o} - \frac{1}{v_o}$
Substitute values:
$\frac{1}{u_o} = \frac{1}{0.50} - \frac{1}{13.5} = 2 - \frac{2}{27}$
$\frac{1}{u_o} = \frac{54}{27} - \frac{2}{27} = \frac{52}{27}$
$u_o = \frac{27}{52}$ cm

Step 3: Lateral magnification by objective
$m_o = \frac{v_o}{|u_o|}$
Substitute values:
$m_o = \frac{13.5}{\frac{27}{52}} = 13.5 \times \frac{52}{27}$
$13.5 = \frac{27}{2}$
$m_o = \frac{27}{2} \times \frac{52}{27} = \frac{52}{2} = 26$

Step 4: Angular magnification by eyepiece (image at infinity)
$m_e = \frac{D}{f_e}$
Substitute values:
$m_e = \frac{25}{2.5} = 10$

Step 5: Total magnification
$m = m_o \times m_e$
Substitute values:
$m = 26 \times 10 = 260$
Final answer: $m = 260$