Question 251:
easyA lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since:
(2021)
A large aperture objective collects more light, which increases the light gathering power, making faint distant objects visible. It also inherently increases the resolving power of the telescope, allowing finer details to be resolved. Therefore, all the given statements are correct.
Question 252:
easyAn astronomical telescope has objective and eyepiece of focal length 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance:
(2016 – I)
For the objective, $u_o = -200 cm$, $f_o = +40 cm$. Using $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$, we get $\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200}$, so $v_o = 50 cm$. In normal adjustment, the intermediate image forms at the focus of the eyepiece, so $u_e = f_e = 4 cm$. The required separation is $L = v_o + f_e = 50 + 4 = 54 cm$.
Question 253:
easyFor a prism its refractive index is $\cot A/2$ then minimum angle of deviation is:
(1999)
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 numerators: $\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$. This gives $A + \delta_m = 180^\circ - A$, so the minimum angle of deviation is $\delta_m = 180^\circ - 2A$.
Question 254:
easyThere is a prism with refractive index equal to $\sqrt{2}$ and the refractive angle equal to $30^\circ$. One of the refractive surface of the prism is polished. A beam of monochromatic light will retrace its path if its angle of incidence over the refracting surface of the prism is
(1992)
For the ray to retrace its path, it must strike the silvered surface normally, so $r_2 = 0$. From $r_1 + r_2 = A$, we have $r_1 = 30^\circ$. Using Snell's law at the first surface, $1 \cdot \sin(i) = \mu \sin(r_1) = \sqrt{2} \sin(30^\circ) = \sqrt{2}(1/2) = 1/\sqrt{2}$. Therefore, the angle of incidence is $i = 45^\circ$.
Question 255:
easyA ray is incident at an angle of incidence i on one surface of a prism of small angle A and emerge normally from opposite surface. If the refractive index of the material of prism is $\mu$, the angle of incidence i is nearly equal to:
(1989)
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 the angle of incidence $i = \mu A$.
Question 256:
easyPick the wrong answer in the context with rainbow.
(2019)
An observer can only see a rainbow when looking away from the sun, so the sun must be behind them. Thus, the statement that the observer's front is towards the sun is incorrect.
Question 257:
easyRainbow is formed due to:
(2000)
A rainbow is formed due to the dispersion of sunlight by water droplets, followed by total internal reflection and refraction. Therefore, total internal reflection and dispersion is the correct combined effect.
Question 258:
easyThe blue colour of the sky is due to the phenomenon of
(1994)
The blue colour of the sky is due to Rayleigh scattering. Since blue light has a shorter wavelength, it is scattered much more strongly by atmospheric particles than longer wavelengths like red.
Question 259:
easyA biconvex lens has radii of curvature, $20 cm$ each. If the refractive index of the material of the lens is $1.5$, the power of the lens is:
(2022)
Using Lens Maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. $\frac{1}{f} = (1.5 - 1)(\frac{1}{20} - \frac{1}{-20}) = 0.5 \times \frac{2}{20} = \frac{1}{20} cm^{-1}$. So focal length $f = 20 cm = 0.2 m$. Power $P = \frac{1}{f(m)} = \frac{1}{0.2} = +5D$.
Question 260:
easyThe power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. Then the refractive index of the material of the lens is,
(2020-Covid)
Power $P = 10 D$, so focal length $f = 1/10 m = 10 cm$. Using the lens maker's formula: $\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. $\frac{1}{10} = (\mu - 1)(\frac{1}{10} - \frac{1}{-10}) = (\mu - 1)(\frac{2}{10})$. This gives $1 = 2(\mu - 1)$, resulting in $\mu = 3/2$.