﻿ Two thin lenses of power +4D and -2D are in contact. What is the focal length of the combination? from Physics Ray Optics and Optical Instruments Class 12 Mizoram Board

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Two thin lenses of power +4D and -2D are in contact. What is the focal length of the combination?

Power of first lens = +4 D
Power of second lens = -2 D
Power of the combination of lens,

Since focal length,
Focal length of the lens,

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Show that the least possible distance between an object and its real image in a convex lens is 4f where f is the focal length of the lens. Suppose I is the real image of an object O. Let d be the distance between them. If the image distance is x, the object distance will be (d – x).
Thus,    u = – (d – x) and v = + x
Sustituting in the lens formula we have or, or,  For a real image, the value of x must be real, i.e., the roots of the above equation must be real. This is possible if
d2 ≥ 4fd
or,    d ≥ 4f
Hence, 4f is the minimum distance between the object and its real image formed by a convex lens.

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An object 0.5 cm high is placed 30 cm from a convex mirror whose focal length is 20 cm. Find the position, size and nature of the image.

We have,
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm

We know, using the mirror formula,

$⇒$

The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror.

Now, size if the object, O = 0.5 cm
sixe of the image = I
Magnification,
$⇒$

Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size

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A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change in its focal length when it is immersed in water of refractive index 4/3.

Given, a convex lens.
Refractive index of lens, ${}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}}$  = 1.5
Focal length of lens in air, fa = 18
Refractive index of water,

For the lens in air,

When the lens is immersed in water

Thus,

Hence, focal length changes from 18 to 32.

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An object 0.2 cm high is placed 15 cm from a concave mirror length 5 cm. Find the position and size of the image.

We have,

Using formula,

The image is formed at a distance of 7.5 cm in front of mirror.
Now,
Magnification,

where,
I is the image size and,

O is the object size.

The negative sign indicates that the image is inverted and is diminished in size.

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A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes a king fisher bird diving vertically towards water at the rate 9 m/s vertically above it. If the refractive index of water is 4/3, find the actual velocity of the dive of the bird.

If at any instant, the fish is at a depth ‘x’ below water surface while the bird at a height y above the surface, then the apparent height of the bird from the surface as seen by the fish will be given by  or Apparent height = μy
So, the total apparent distance of the bird as seen by the fish in water will be h = x + μy
or, or, or, 2778 Views