It is a point on the principal axis at which a beam of light incident parallel to the principal axis converges after reflection from the concave mirror.

Focal length = \(\frac{1}{2}\) x Radius of curvature = \(\frac{1}{2}\) x 20 = 10 cm

Concave mirror.

The field of view of a convex mirror is wider than that of concave mirror and convex mirror always produces erect image of object, so we prefer convex mirror as rear view mirror for vehicles.

Given, radius of curvature, R = 32 cm

As we know, focal length, f = \(\frac{\text{radius of curvature}}{2}\) = 16 cm

As the image is real, so magnification m must be negative.

m = \(\frac{h'}{h}=-\frac{v}{u}\) = -3

∴ v = 3 u

Given u = - 10 cm

∴ u = 3 x (- 10) = - 30 cm.

Thus the image is located at a distance of 30 cm from the mirror on the object side of the mirror.

Light bends towards the normal on entering water. It happens because water is an optically denser medium than air. When light travels from a rarer medium to denser medium, it bends towards the normal.

Refractive index of glass, n_g = 1.50

Speed of light in vacuum, c = 3 x 10⁸ ms^-1

Speed of light in glass = \(\frac{c}{n_g}\) = \(\frac{3×10^8}{1.5}\) = 2 x 10⁸ ms^-1

From Table,

• Diamond has highest refractive index (= 2.42), so it has largest optical density.
• Air has lowest refractive index (= 1.0003), so it has lowest optical density.

For kerosene, n = 1.44

For turpentine oil, n = 1.47

For water, n = 1.33

As water has lowest refractive index, so light travels fastest in this optically rarer medium than kerosene and turpentine oil.

Refractive index of any medium w.r.t. another indicates the extent to which light bends, when it enters from first medium to the given medium.

The given value of refractive index also states that speed of light in diamond is 1/2.42 times to the speed of light in vacuum.

One dioptre is the SI unit of power of lens, whose focal length is 1 m.

Here, v = +50 cm,                 ….[v is positive for real image]

As the real image is of the same size as the object, so

m = \(\frac{h'}{h}=-\frac{v}{u}\) = -1

∴ u = - v = - 50 cm.

Now \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{50}-\frac{1}{-50}\) = \(\frac{1}{50}+\frac{1}{50}\) = \(\frac{2}{50}=\frac{1}{25}\)

∴ f = 25 cm = 0.25 m

P = \(\frac{1}{f}\) = \(\frac{1}{0.25}\) = +4 D

As the focal length of a concave lens is negative, so f = -2 m

P = \(\frac{1}{f}\) = \(\frac{1}{2}\)  = -0.5 D

(d) Clay

Clay can never be transparent, so it cannot be used to make lens.

(d) Between the pole of the mirror and its principal focus.

If the object is placed between pole of the mirror and its principal focus virtual, erect and magnified image will be formed.

(b) At twice the focal length

To get the real image of the size of the object, it should be placed at twice the focal length of a convex lens.

(a) both concave.

The focal length is taken as negative for both concave mirror and concave lens.

(d) either plane or convex.

Plane mirrors and convex mirrors always form the erect images.

(c) A convex lens of focal length 5 cm.

Convex lens is used as magnifying glass.

As, focal length, f = \(\frac{1}{P}\)

Convex lens of focal length 50 cm has power 2 D,

whereas convex lens of focal length 5 cm has power 20 D.

Hence for, better performance its focal length should be small.

The object should be kept at a distance less than 15 cm from the mirror.

The image is virtual and erect.

The image is larger than the object.

(a) Concave mirror to get powerful beam of light.

(b) Convex mirror to get larger field of view and erect image.

(c) Concave mirror, as it can converge light rays of the sun on a small area.

The lens will produce a complete image, but the image will have lower intensity and brightness because in the lower part of the lens the rays are blocked.

Given, object size, h_o =5 cm; Image size, h_i =?

Object distance, u = - 25 cm; Image distance, v =?

Focal length, f = 10 cm

From lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Or \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) = \(\frac{1}{10}+\frac{1}{-25}\) = \(\frac{3}{50}\)

Or v = \(\frac{50}{3}\) = 16.67 cm from the lens

Size of image,

h_o = h_i x \(\frac{v}{u}\) = 5 x \(\frac{50}{3×(-25)}\) = - 3.33 cm    

The image is real in nature.

Given, focal length, f = - 15 cm

Image distance, v = - 10 cm

By lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

∴ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\) = \(\frac{1}{-10}-\frac{1}{-15}\) = \(-\frac{1}{30}\)

u = - 30 cm from the lens.

Thus, the object is placed at 30 cm from the concave lens.

Given, object distance, u = - 10 cm

Focal length, f = 15 cm

By mirror formula, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) = \(\frac{1}{15}-\frac{1}{-10}\) = \(\frac{5}{30}\) = \(\frac{1}{6}\)

v = 6 cm from the mirror

Image is formed 6 cm behind the mirror.

So, image is virtual and erect.

If magnification is 1, the image size is same as that of object size.

Since, magnification is positive, the image is erect and virtual.

Given : object size, h_o = 5 cm, Object distance, u = - 20 cm,

Radius of curvature = 30 cm

Focal length, f = \(\frac{30}{2}\) = 15 cm

Position of image (by mirror formula)

\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)  or \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) = \(\frac{1}{15}-\frac{1}{-20}\) = \(\frac{7}{60}\) 

v = \(\frac{60}{7}\) = 8.57 cm

Image is formed at 8.57 cm behind the mirror.

∴ Magnification, m = \(\frac{h_i}{h_o}\) = \(-\frac{v}{u}\) 

= \(-\frac{60}{-20×7}\) = \(\frac{h_i}{5}\)

∴ h_i = \(-\frac{5×60}{-20×7}\) = 2.14 cm

Image is diminished, erect and hence virtual.

Given: object size, h₀ = 7 cm, Object distance, u = - 27 cm

Focal length, f = - 18 cm

By mirror formula, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) = \(\frac{1}{-18}-\frac{1}{-27}\) = \(-\frac{1}{54}\) 

Screen should be placed at 54 cm in front of the mirror.

Now, \(\frac{h_i}{h_o}\) = \(-\frac{v}{u}\) 

\(\frac{h_i}{7}\) = \(-\frac{-54}{-27}\)  = -2

h_i = 7 x -2 = -14 cm

So, image is double the size of object. Also, image is real and inverted (since, h_i is negative).

Focal length, f = \(\frac{1}{Power}\) = \(\frac{1}{-2}\) - 0.5 m

Since, f is negative, lens is concave.

Given : power of lens, P = +1.5 D

It means the lens is convex

As power, P = \(\frac{1}{f(in\,m)}\)

f = \(\frac{1}{P}\) = \(\frac{1}{+1.5}\) = 0.66 m

f = 0.66 × 100 = 66 cm

[It is a converging lens because its focal length is positive.]