Chapter 9: Light Reflection and Refraction

Ans: To obtain an erect image in a concave mirror the object should be placed between Focus and the Optic centre.

Here, the focal length of the concave mirror is given as $15cm$.

Therefore, the range of distance of the object from the mirror is from $0cm$ to$15cm$.

The nature of the image is virtual.

The image is larger than the object. 

A virtual, erect and magnified image is formed.

Ans: In the headlights of a car, a concave mirror is used. Because in concave mirrors a parallel beam of light is produced if the bulb is placed at the focus.

1. Side/Rear-View Mirror of a Vehicle

Ans: In a side/rear-view mirror of a vehicle , a convex mirror is used. Because when objects are placed in front of the convex mirror, erect and diminished images are formed which gives a wider field of view.

1. Solar Furnace

Ans: In solar furnaces, Concave mirrors are used. They converge sunlight to a point and produce high temperatures because of their converging properties.

Ans:  Yes, the lens produces a complete image of the object with less intensity.

Consider the following two cases:

In the first case the lower half of the lens is covered with black paper. Light rays coming from the object are refracted only from the upper half and the image is formed, whereas in the lower half the light rays are blocked.

In the second case the upper half of the lens is covered with black paper. Light rays coming from the object are refracted only from the lower half and the image is formed, whereas in the upper half the light rays are blocked.

Therefore, change in intensity of the image is observed i.e., the intensity of the image is less and the complete image is formed. 

Ans: It is given that,

Height of the object,${{h}_{o}}=5cm$

Distance of object in front of lens,$u=-25cm$

Distance of image from lens,$v=?$

Focal length of the lens,$f=+10cm$ 

From lens formula:$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{25}$ 

$\Rightarrow \frac{1}{v}=\frac{25-10}{250}$ 

$\Rightarrow \frac{1}{v}=\frac{15}{250}$ 

$\Rightarrow v=\frac{250}{15}$ 

$\Rightarrow v=16.66cm$ 

The positive value of $v$ indicates that the image and the object are on opposite sides.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

Height of the image, ${{h}_{i}}=?$ 

$\Rightarrow m=\frac{{{h}_{i}}}{5}=-\frac{16.66}{25}=-0.66$ 

$\Rightarrow {{h}_{i}}=-0.66\times 5$ 

$\Rightarrow {{h}_{i}}=-3.3cm$

As the magnification is $-0.66$, the negative sign indicates that the object is inverted and less than $1$ indicates that the image is smaller than the object.

Therefore, the position of the image is $16.66cm$ from the lens. The height of the object is $3.3cm$. The nature of the image is real, inverted and diminished.

Ans: It is given that, 

Focal length of the lens,$f=-15cm$ 

Distance of image from lens,$v=-10cm$

Distance of object in front of lens,$u=?$

From lens formula:$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$ 

$\Rightarrow \frac{1}{u}=\frac{1}{(-10)}-\frac{1}{(-15)}$ 

$\Rightarrow \frac{1}{u}=\frac{-1}{10}+\frac{1}{15}$ 

$\Rightarrow \frac{1}{u}=\frac{-5}{150}$ 

$\Rightarrow u=-\frac{150}{5}$ 

$\Rightarrow u=-30cm$ 

Thus, the object is at a distance of $30cm$ from the lens.

Ans: It is given that,

Focal length of the convex mirror,$f=+15cm$ 

Distance of object in front of convex mirror,$u=-10cm$

Distance of image from convex mirror,$v=?$

From mirror formula:$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{(-10)}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}+\frac{1}{10}$ 

$\Rightarrow \frac{1}{v}=\frac{25}{150}$ 

$\Rightarrow v=\frac{150}{25}$ 

$\Rightarrow v=6cm$ 

The positive value of $v$ indicates that the image is formed behind the mirror and virtual.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

$\Rightarrow m=\frac{-6}{-10}=+0.6$ 

The positive magnification indicates that the image is erect. As magnification is less than $1$ it indicates that the image is smaller than the object.

Therefore, the position of image is $6cm$ behind the mirror. Nature of image is virtual, erect and diminished.

Ans: It is known that,

Magnification produced by a plane mirror is,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

where,

${{h}_{i}}$ is the height of the image

${{h}_{o}}$ is the height of the object

$u$ is the distance of the object in front of lens

$v$ is the distance of image from lens

Magnification produced by a plane mirror is $+1$ means that ${{h}_{i}}={{h}_{o}}$ that is the size of the image is the same as the size of the object. The positive size indicates that the image is erect.

Therefore, magnification equal to $+1$ means that the size of the image is the same as object and erect.

Ans: It is given that,

Distance of object in front of the mirror,$u=-20cm$

Distance of image from the mirror,$v=?$

The radius of curvature of the mirror,$R=30cm$

The focal length of the mirror,$f=?$ 

It is known that,

Radius of curvature is equal to twice the focal length.

$\Rightarrow R=2f$ 

$\Rightarrow 30=2f$ 

$f=\frac{30}{2}=15cm$ 

From mirror formula:$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{(-20)}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}+\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{20+15}{300}$ 

$\Rightarrow \frac{1}{v}=\frac{35}{300}$ 

$\Rightarrow v=\frac{60}{7}$ 

$\Rightarrow v=8.57cm$ 

The positive value of $v$ indicates that the image is formed behind the mirror.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

Height of the object, ${{h}_{o}}=5cm$

Height of the image, ${{h}_{i}}=?$ 

$\Rightarrow {{h}_{i}}=-\frac{v}{u}\times {{h}_{o}}$ 

$\Rightarrow {{h}_{i}}=-\frac{8.57}{(-20)}\times 5$ 

$\Rightarrow {{h}_{i}}=2.14cm$

Therefore, the image is formed at a distance of \[8.57cm\]behind the mirror. The nature of the image is virtual, erect and diminished.

Ans: It is given that,

Distance of object in front of mirror,$u=-27cm$

Distance of image from the mirror,$v=?$

Focal length of the mirror,$f=-18cm$ 

From mirror formula:$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{-18}-\frac{1}{(-27)}$ 

$\Rightarrow \frac{1}{v}=-\frac{1}{18}+\frac{1}{27}$

$\Rightarrow \frac{1}{v}=\frac{-3+2}{54}$ 

$\Rightarrow \frac{1}{v}=-\frac{1}{54}$ 

$\Rightarrow v=-54cm$ 

The negative value of $v$ indicates that the screen should be placed at a distance of  $54cm$ in front of the mirror and the image is real.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

Height of the object, ${{h}_{o}}=7cm$

Height of the image, ${{h}_{i}}=?$ 

$\Rightarrow {{h}_{i}}=-\frac{v}{u}\times {{h}_{o}}$ 

$\Rightarrow {{h}_{i}}=-\frac{-54}{(-27)}\times 7$

$\Rightarrow {{h}_{i}}=-14cm$

The height of image is \[14cm\].

Therefore, the image is formed at a distance of \[54cm\]in front of the mirror. Nature of image is real, inverted and enlarged.

Ans: It is given that,

Power of a lens,$P=-2.0D$ 

Focal length of a lens, $f=?$ 

Power of a lens,$P=\frac{1}{f(metres)}$ 

$\Rightarrow -2=\frac{1}{f}$ 

$\Rightarrow f=-\frac{1}{2}=-0.5m$ 

Negative $f$ indicates concave lens.

Therefore, the focal length of lens is $f=-0.5m$and the lens is concave.