Advertisements
Advertisements
Question
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
Solution
`m_e = D/f_e = 25/5 = 5`
m = m0 × me
`m_0 = m/m_e = -30/5 = -6`
`m_0 = v_0/u_o, -6 = v_0/u_o`
v0 = −6ue
`1/f_e = 1/v_0 - 1/u_0`
`= 1/1.25 = 1/(-6u_0) - 1/u_0 = (-1-6)/(6u_0) = -7/(6u_0)`
`= u_0 = (-7xx1.25)/6`
= −1.46 cm
APPEARS IN
RELATED QUESTIONS
State the condition under which a large magnification can be achieved in an astronomical telescope.
Draw a ray diagram showing the image formation of a distant object by a refracting telescope ?
Describe briefly the two main limitations and explain how far these can be minimized in a reflecting telescope ?
Define its magnifying power and write the expression for it?
Define magnifying power of a telescope. Write its expression.
An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.
A Galilean telescope is 27 cm long when focussed to form an image at infinity. If the objective has a focal length of 30 cm, what is the focal length of the eyepiece?
Draw a labelled ray diagram showing the formation of an image by a refracting telescope when the final image lies at infinity.
Draw a ray diagram of a refracting astronomical telescope when final image is formed at infinity. Also write the expression for its angular magnification (magnifying power).
Draw a ray diagram of Astronomical Telescope for the final image formed at infinity
