Question 241:
easyTwo thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combinations is:
(2008)
Power of the combination is the sum of individual powers. $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_1 + f_2}{f_1 f_2}$.
Question 242:
easyA convex lens and a concave lens, each having same focal length of $25 cm$, are put in contact to form a combination of lenses. The power in diopters of the combination is:
(2006)
For the convex lens, $f_1 = +25 cm$, and for the concave lens, $f_2 = -25 cm$. Equivalent power is $P = P_1 + P_2 = \frac{100}{f_1} + \frac{100}{f_2}$. $P = \frac{100}{25} - \frac{100}{25} = 0$.
Question 243:
easyFor a plano convex lens ($\mu = 1.5$) has radius of curvature $10 cm$. It is silvered on its plane surface. Find focal length after silvering:
(2000)
The equivalent power of the silvered lens is $P = 2P_L + P_M$. The power of the plane mirror is $P_M = 0$. The power of the lens is $P_L = (\mu - 1)\left(\frac{1}{R}right) = (1.5 - 1)\left(\frac{1}{10}right) = \frac{0.5}{10} cm^{-1}$. Equivalent power $P = 2 \times \frac{0.5}{10} = \frac{1}{10} cm^{-1}$, giving a focal length magnitude of $F = 10 cm$.
Question 244:
easyIf a convex lens of focal length $80 cm$ and a concave lens of focal length $50 cm$ are combined together, what will be their resulting power?
(1996)
The powers of the lenses are $P_1 = \frac{100}{80} = 1.25 D$ and $P_2 = \frac{100}{-50} = -2.0 D$. The resulting power of the combination is $P = P_1 + P_2$. $P = 1.25 - 2.0 = -0.75 D$.
Question 245:
easyA ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is m, then the angle of incidence is nearly equal to :
(2020)
Since the ray emerges normally, the angle of emergence $e = 0$, which implies $r_2 = 0$. For a prism, $r_1 + r_2 = A$, so $r_1 = A$. Using Snell's law for small angles, $i = \mu r_1$, which gives $i = \mu A$.
Question 246:
easyThe refractive index of the material of a prism is $\sqrt{2}$ and the angle of the prism is $30^\circ$. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is:
(2018)
To retrace its path, the ray must strike the silvered surface normally, meaning $r_2 = 0$. From $r_1 + r_2 = A$, we get $r_1 = A = 30^\circ$. Applying Snell's law at the first surface: $1 \cdot \sin(i) = \mu \sin(r_1) = \sqrt{2} \sin(30^\circ) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Therefore, the angle of incidence is $i = 45^\circ$.
Question 247:
easyThe angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ$. The angle of prism is $60^\circ$. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:
(2016 – I)
At minimum deviation, angle of emergence equals angle of incidence, so $i = e = 45^\circ$. Minimum deviation $\delta_m = i + e - A = 45^\circ + 45^\circ - 60^\circ = 30^\circ$. Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
Question 248:
easyThe refracting angle of a prism is A, and refractive index of the material of the prism is $\cot(A/2)$. The angle of minimum deviation is:
(2015)
Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$. Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin(90^\circ - A/2)}{\sin(A/2)}$. Equating the numerators, we get $\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$. This simplifies to $A + \delta_m = 180^\circ - A$, so the angle of minimum deviation is $\delta_m = 180^\circ - 2A$.
Question 249:
easyThe angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index $\mu$, of the prism is:
(2014)
For the ray to retrace its path, it must strike the silvered surface normally ($r_2 = 0$), so $r_1 = A$. Using Snell's law at the first surface: $\sin(i) = \mu \sin(r_1)$. Substituting $i = 2A$ and $r_1 = A$, we get $\sin(2A) = \mu \sin(A)$. Since $\sin(2A) = 2\sin(A)\cos(A)$, we find $\mu = 2\cos A$.
Question 250:
moderateIn compound microscope the magnification is 95, and the distance of object from objective lens 1/3.8 cm and focal length of objective is 1/4 cm. What is the magnification of eye pieces when final image is formed at least distance of distinct vision:
(1999)
Objective magnification is $m_o = \frac{f_o}{f_o + u_o}$. Substituting $f_o = 0.25$ and $u_o = -1/3.8$, $m_o = \frac{1/4}{1/4 - 1/3.8} = \frac{1/4}{(3.8 - 4)/15.2} = -19$. Total magnification is $M = m_o \times m_e$, so $95 = 19 \times m_e$ (using magnitudes). This gives the magnification of the eyepiece as $m_e = 5$.