Compound Microscope

COMPOUND MICROSCOPE AND TELESCOPE Compound Microscope Objective lens: Near the object. Eyepiece: Near the eye. U_{0}*z The distance of the object from the objective lens V 0 ^ 2 The distance of the image formed by the objective lens in relation to itself u_{c}*t The distance of the image formed by the objective lens in relation to the eyepiece Object lens f_{p} Eye lens l_{e} u_{0} v_{0} 2f_{0} overline t_{0} f_{0} 2f_{n} h ^ prime prime v_{e} The image produced by the objective lens serves as the object for the eyepiece, and it's crucial for magnification that this image falls between the focus and the optical center of the eyepiece. Using the lens, the visual angle becomes, Without using the lens, the visual angle becomes, Therefore, the angular magnification or the magnifying power becomes, tan theta = theta \approx (h')/u_{e} MP = (- h')/(u_{e} * h') * D MP = (- theta)/theta_{0} theta = (h')/d Given that h and h represent the height of the object and image with respect to the objective lens. represents the lateral magnification for the objective lens, and D/u_{e} denotes the magnifying power of the eyepiece. (h ^ r)/h Alternatively, the lateral magnification for the objective lens can be expressed as: b^ prime h = -v p -u 0 Rightarrow (h')/h = v_{p}/a_{0} So, the angular magnification for the compound microscope is: M.P.= - v_{0}/u_{0} * D/u_{v} Minimum Magnifying Power Normal State Applying thin lens formula for eyepiece, [ gamma*u =-u e^ prime ] v = - v_{e} and f = f_{e} ] 1/u_{e} = [1/v_{0} + 1/t_{e}] - 1/v_{v} + 1/u_{v} = 1/t_{v} For normal state, v e = infty Rightarrow u_{e} = f_{e} Therefore, the magnifying power becomes, Minimum f.P . = - v_{0}/u_{0} * D/t_{u} u_{e} = f_{e} M.P.= - v_{0}/u_{0} * D/u_{0} Minimum Magnifying Power: Normal State △ABC and AADE are similar triangle. Thus, we can write: (DE)/(BC) Rightarrow AE AC Rightarrow h^ prime h Rightarrow L f 0 1. represents the distance between the second focus of the objective and the first focus of the eyepiece. Hence, the magnification power is obtained as MP = - v_{0}/u_{0} * D/t_{0} = - [(h')/h] * D/f_{e} = - L/t_{0} * D/t_{w} Minimum M. P * = - L/f_{0} * D/t_{0} To enhance the magnification power, both the focal lengths of the objective lens and the eyepiece should be minimized. Maximum Magnifying Power Applying thin lens formula for eyepiece: [vu-v-v-Dand f Therefore, the magnifying power becomes, Maximum M.P.= - v_{o}/v_{0} * (1 + D/I_{e}) 1/(- D) + 1/u_{e} = 1/t_{e} 1/u_{w} = 1/epsilon_{o} + 1/D v_{e} = D M.P . =- v 0 u 0 * D u e Rightarrow M.P.= - (v_{0}*D)/u_{0} * (1/t_{e} + 1/D) Compound Microscope z_{2} epsilon_{0} epsilon_{V} h ^ n | Real Set Up Eye Lens 2 5 Object Lens z_{0} Simple Microscope vs Compound Microscope Simple Microscope Compound Microscope (M.P.) min = D/f (M.P.) min = - v_{0}/u_{0} * D/epsilon_{w} = - L/epsilon_{0} * D/epsilon_{w} (M.P,) min = - v_{0}/u_{0} * D/t_{o} = - L/t_{0} * D/t_{o} (M.P.) mitx = - v_{0}/u_{0} * (1 + D/t_{w}) y/u_{0} u_{p} C * Telescope Telescope Refracting Telescope Lenses are used Reflecting Telescope Converging mirror Are used Astronomical Telescope Terrestrial Telescope Astronomical Telescope Magnifying power The telescope serves to enhance the angular size of remote objects, such as stars or planets. Objective: In the direction of the object (with a longer focal length and larger aperture). Eyepiece: Near the eye (with a shorter focal length and smaller aperture). Due to the incident ray on the objective lens being a parallel beam of light but not parallel to the principal axis, the image produced by the objective lens will appear at the focal plane of the object. Objective lens Eyepiece lens The angle subtended by the object ( theta ) at the optical center of the objective The angle subtended by the final image (0) at the optical center of the eyepiece Therefore the magnifying power of telescope is given by P . = theta/theta_{0} = 1/u_{e} * t_{0}/1 y/t_{0} (The-ve sign signifies0 the final image is inverted w.r.t the object) M.P . = - f_{e}/u_{e} c_{6} \mathcal{J}*a_{g} Objective lens Magnifying Power: Minimum Utilizing the lens formula for the eyepiece: Hence the magnifying power become For minimum magnifying power final image is formed at infinity i.e. v_{e} -> ∞ Therefore minimum magnifying power is. M* P min =- t n t p M.P.= - t_{o}/u_{e} = - f_{0} * [1/v_{c} + 1/t_{e}] 1/u_{e} = 1/f_{e} + 1/v_{e} 1/(- v * e) + 1/u_{e} = 1/f_{e} u_{e} = f_{e} I_{v} = - v_{e} * I u = - u_{e} f_{a} f_{a} Objective lens u_{0} f * Eyepiece lens Magnifying Power: Maximum We have: In this case final image formed at the least distance of district vision, thus. Therefore the magnifying power become. MP = - r_{o}/u_{r} 1/u_{v} = 1/t_{0} + 1/D M* P max =-f o ( 1 t o + 1 D ) 1/u_{v} = 1/t_{o} + 1/v_{e} |v_{e}| = D To increase the magnifying power, the focal length of objective lens should be large and that of the eyepiece should be small. f_{0} theta_{u}*C f_{a} g_{0} u_{e} r_{e} Eyepiece lens Objective lens D Tube length The length of the telescope tube is determined by L = f_{o} + u_{e} Maximum value of tube length Maximum possible value of ue is Maximum value of L is, L max =f o +f e U max =f e t_{0} L 0 f_{0} u_{n} Objective Eyepiece lens Minimum value of tube length: In this case the required condition is v_{e} = D Therefore, u = - u_{e} v = - v_{o} = - D f = f_{e} applying then lens formula for eyepiece we get. - 1/D + 1/u_{e} = 1/t_{e} 1/u_{e} = 1/t_{0} + 1/D u_{e} = (t_{e}*D)/(t_{e} + D) Hence minimum value of L is, In order to capture the maximum amount of light from distant objects, the aperture of the objective lens should be sizable. L min =f o + f e D f w +D Tube length (L) C For decent intensity of light, D o >D ef In general case, f_{o} >f e f_{o} > f_{e} R_{0} For minimum magnifying power we have: Thus, tube length becomes: In general case, f_{o} >f e Therefore, L cong f 0 Formula of minimum magnifying power is approximately written as follows: MP = - t_{0}/t_{w} \approx - h/t_{w} , Rightarrow adjustment L=f o +u e Rightarrow L = f_{o} + f_{e} u_{e} = f_{e} f_{e} u_{n} f_{w} Objective lens L Byeplece lens Ex. An astronomical telescope has an objective lens of focal length 40 cm and eyepiece lens of focal length 10 cm. Find the minimum magnifying power and tube length of telescope. Sol Given: The focal length of the objective lens: f_{o} = 40cm The focal length of the eyepiece: f_{c} = 10c 40 epsilon_{e} = 12cos z Minimum magnifying power is given by, The tube length of the telescope is given by, M.P min =-4 M* P min =- t 0 t e = -40 +10 =-4 L = f_{o} + f_{e} = 40 + 10 = 50cm L = 50cm Ex. Aperture of an objective and eyepiece lens are D_{0} and D_{8} respectively of object I splaced at infinity and its final image is formed at infinity then find the minimum magnifying power of an astronomical telescope. Sol Since the and the final image both are formed at infinity the ray diagram for the problem will be as shown in the adjacent figure and from the figure we can write. f_{o}/D_{o} = f_{e}/D_{e} D O_{2} Objective lenn Kyrpiece The minimum magnifying power of an astronomical telescope becomes. P min =- epsilon o t n = -D 0 D n M* P min =- D o D e Terrestrial Telescope Employing an astronomical telescope for observation. Outer-space object Terrestrial Object The astronomical telescope produces an inverted image, thus making it unsuitable for viewing objects on Earth. A convex lens exhibits a characteristic: When an object is positioned at twice the focal length (2f), an inverted image is created at twice the focal length on the opposite side of the object. Therefore, in the configuration of an astronomical telescope, a convex lens is placed between the objective and the eyepiece, causing it to produce an erect image. Objective lens Erecting Kyrpiece The magnification capability and the length of the telescope tube will be: M.P L=f+4f+Uc [f is the focal length of the erecting lens) Used to see erect images of distant earthly objects. Astronomical telescope Inverted image Terrestrial telescope Erect image Length is much larger than the astronomical telescope. Extra reflection at erecting lens result in decrease in intensity of final image.