A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be:
(2016 – II)
To see objects at infinity, the correcting lens must form an image at the person's far point, so $u = -\infty$ and $v = -400 cm = -4 m$. Power $P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-4} - \frac{1}{-\infty}$. $P = -0.25 D$. A negative power indicates a concave lens.
For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea of eye lens can be estimated to be:
(2013)
The total converging power of the eye is $P = P_1 + P_2 = 40 D + 20 D = 60 D$. The focal length is $f = \frac{1}{P} = \frac{1}{60} m = \frac{100}{60} cm = \frac{5}{3} cm$. The distance between the retina and the cornea is equivalent to this focal length, which is $\approx 1.67 cm$.
If the focal length of objective lens is increased then magnifying power of:
(2014)
The magnifying power of a microscope is inversely proportional to the focal length of its objective ($M \propto 1/f_o$). The magnifying power of a telescope is directly proportional to the focal length of its objective ($M = f_o/f_e$). Therefore, increasing $f_o$ decreases microscope magnification and increases telescope magnification.
In an astronomical telescope in normal adjustment a straight black line of length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is:
(2015 Pre)
Magnification $m = \frac{f_o}{f_e}$. The objective acts as an object for the eyepiece at distance $u = -(f_o + f_e)$. Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_e}$, we get $v = \frac{f_e(f_o+f_e)}{f_o}$. Magnification of this image is $\frac{I}{L} = \left|\frac{v}{u}\right| = \frac{f_e}{f_o}$, so the telescope magnification is $\frac{L}{I}$.
The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are:
(2012 Pre)
Magnifying power $M = \frac{f_o}{f_e} = 9$, which gives $f_o = 9f_e$. Distance between lenses in normal adjustment is $L = f_o + f_e = 20 cm$. Substituting $f_o$, we get $9f_e + f_e = 20$, so $10f_e = 20 \Rightarrow f_e = 2 cm$. Then $f_o = 18 cm$.
A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the Sun is $1.39 \times 10^9 m$ and its mean distance from the earth is $1.5 \times 10^{11} m$. What is the diameter of the Sun’s image on the paper?
(2008)
The angle subtended by the Sun at the lens is $\alpha = \frac{D}{d} = \frac{1.39 \times 10^9}{1.5 \times 10^{11}} = 0.926 \times 10^{-2} rad$. The image is formed at the focus, so the angle subtended by the image is $\alpha = \frac{d'}{f}$. Thus, the image diameter $d' = \alpha f = 0.926 \times 10^{-2} \times 0.1 = 9.26 \times 10^{-4} m$.
Four lenses of focal length $\pm 15 cm$ and $\pm 150 cm$ are available for making a telescope. To produce the largest magnification, the focal length of the eyepiece should be
(1994)
Magnification of a telescope is given by $M = \frac{f_o}{f_e}$. For the largest magnification, the objective should have the maximum focal length and the eyepiece should have the minimum focal length magnitude. For an astronomical telescope, both lenses are convex, so the eyepiece should be $+15 cm$.
A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is $5000 \AA$, is of the order of:
(2004)
Resolving power limit $\theta = \frac{1.22 \lambda}{D} = \frac{1.22 \times 5000 \times 10^{-10}}{0.1} = 6.1 \times 10^{-6} rad$. The minimum linear distance is $x = r \theta = 1000 \times 6.1 \times 10^{-6} = 6.1 \times 10^{-3} m = 6.1 mm$, which is of the order of $5 mm$.
A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle $\theta$, the spot of the light is found to move through a distance y on the scale. The angle $\theta$ is given by:
(2017-Delhi)
When a plane mirror is rotated by an angle $\theta$, the reflected ray rotates by an angle $2\theta$. For a screen at distance $x$, the displacement of the spot is $y = x \tan(2\theta)$. For small angles, $\tan(2\theta) \approx 2\theta$, so $y = x(2\theta)$, which yields $\theta = \frac{y}{2x}$.