Advertisements
Advertisements
Question
An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) the focal length of the lens, and (ii) the size of the image.
Advertisements
Solution
Object height, h = 4.25 mm = 0.425 cm (1 cm = 10 mm)
Object distance, u =-10 cm
Power, P = +5 D
Focal length, f = ?
Image distance, v = ?
Image height, h' = ?
Power, P `=1/f`
`f=1/p=1/5=0.2m =20cm `
Using the lens formula, we get:
`1/v-1/u=1/f`
`1/v-1/-10=1/20`
`1/v+1/10=1/20`
`1/v=1/20-1/10`
`1/v=-1/20`
∴ v=-20cm
Now, magnification, m = `(h')/h=v/u`
Substituting the values in the above equation, we get:
`(h')/0.425=(-20)/-10`
`h'=2X0.425=0.85cm=8.5mm`
Thus, the image is 8.5 mm long; it is also erect and virtual.
APPEARS IN
RELATED QUESTIONS
The lens mentioned in 6(b) above is of focal length 25cm. Calculate the power of the lens.
A student uses a lens of focal length 40 cm and another of –20 cm. Write the nature and power of each lens
Which type of lens has (a) a positive power, and (b) a negative power?
What is the nature of a lens having a power of + 0.5 D?
What is the nature of a lens whose power is, −4 D?
A convex lens of power 5 D and a concave lens of power 7.5 D are placed in contact with each other. What is the :
(a) power of this combination of lenses?
(b) focal length of this combination of lenses?
What do you understand by the power of a lens? Name one factor on which the power of a lens depends.
A diverging lens has a focal length of 0.10 m. The power of this lens will be:
A symmetric double convex lens in cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4 D, the power a cut-lens will be
Consider three converging lenses L1, L2 and L3 having identical geometrical construction. The index of refraction of L1 and L2 are \[\mu_1 \text{ and } \mu_2\] respectively. The upper half of the lens L3 has a refractive index \[\mu_1\] and the lower half has \[\mu_2\] following figure . A point object O is imaged at O1 by the lens L1 and at O2 by the lens L2placed in same position. If L3 is placed at the same place,
(a) there will be an image at O1
(b) there will be an image at O2.
(c) the only image will form somewhere between O1 and O2
(d) the only image will form away from O2.
A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.
Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?
A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
How is accommodation produced?
A lens forms an upright and diminished image of an object, irrespective of its position. What kind of lens is this? Draw an outline ray diagram to show the formation of the image. State the position and one more characteristic of the image.
If the lens is of focal length 25 cm. Calculate the power of the lens.
The power of the magnifying glass depends on the distance of the magnifying glass from object.
The power of lens depends on the focal length of the lens
Assertion and reasoning type
- Assertion: Myopia is due to the increase in the converging power of eye lens.
- Reason: Myopia can be corrected with the help of concave lens.
