Define the principal focus of a concave mirror.

**Answer:** In case of a concave mirror, all the rays coming from infinity converge at a point after getting reflected from the mirror. This point is called the focus of the mirror.

The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

= 2 x focal length (f)

So, `f = R ÷ 2`

`= 20 cm ÷ 2= 10 cm`

Hence, focal length = 10 cm

Name a mirror that can give an erect and enlarged image of an object.

**Answer:** Concave mirror

Why do we prefer a convex mirror as a rear-view mirror in vehicles?

**Answer:** A convex mirror can show image of a wider area, because of its wide field of view. This enables the driver to see more of traffic coming from behind. Hence, convex mirror is used as rear-view mirror in vehicles.

Find the focal length of a convex mirror whose radius of curvature is 32 cm.

**Answer:** Given, radius of curvature (R) = 32cm and focal length (f) =?

We know that, R = 2f

Hence, 32 cm = 2f

Or, f = 32 cm ÷ 2 = 16 cm

Hence, focal length = 16 cm

A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

**Answer:** Given, the distance of object, u = −10 cm

Let the height of object = h

So, height of image = ÷3h

`m=text(Height of image hâ€™)/text(Height of object h)`

`=text(Distance of image v)/text(Distance of object h)`

Or, `-(3h)/(h)=(v)/(-10 cm)`

Or, `-3=(v)/(-10 cm)`

Or, `v=-30 cm`

Thus, the location of image is at - 30 cm at principal axis before the mirror. The negative sign shows that image is real.

A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

**Answer:** When a ray of light travels from a rarer medium to a denser medium, it bends towards the normal. Air is rarer than water and hence, the ray of light would bend towards the normal.

Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10^{8} m s^{−1}.

**Answer:** Given, refractive index of glass (n_{g}) = 1.50

Speed of light in vaccum (c) = 3.8 × 10^{8}ms^{−1}

Speed of light in glass (v) = ?

We know that n_{g} = c/v

Or, n_{g} = 3.8 × 10^{8} ms^{−1}÷ v

v = 3.8 × 10^{8} ms^{−1}÷ 1.50 = 2 × 10^{8} ms^{−1}

Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

**Answer:** Diamond has the highest optical density, while air has the least optical density.

You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

**Answer:** Among the given matters, water has the least refractive index. Hence, light travels the fastest in water; among the given materials.

The refractive index of diamond is 2.42. What is the meaning of this statement?

**Answer:** We know that refractive index of a medium (n_{m}) = c/v

The above equation shows that the speed of light in vacuum is 2.42 times the speed of light in diamond.

Define 1 dioptre of power of a lens.

**Answer:** The power of a lens with focal length 1 m is called 1 dioptre.

A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

**Answer:** Given, distance of image, v = 50 cm

When an object is kept at the centre of curvature of a convex lens, i.e. at 2F the image formed at the centre of curvature at the other side of lens and the size of image is equal to that of the object.

Thus, the distance of object, u = - 50 cm

We know that;

`1/v=-1/u=1/f`

Or, `(1)/(50)-(1)/(-50)=1/f`

Or, `(2)/(50)=1/f`

Or, `f=25 cm=0.25 m`

Power of lens; `P = 1 ÷ 0.25 = +4 D`

Hence, needle is placed at the centre of curvature, i.e. at 50cm in front of lens

Power of lens = +4D

Find the power of a concave lens of focal length 2 m.

**Answer:** Since it is a concave lens, hence focal length f = -2 m

Power of lens `= 1/f`

Or, `P = 1 ÷ (−2) = −0.50 D`

Thus, power of the given lens `= −0.5 D`

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