Answer

(d) Concave mirror as well as concave lens

Explanation

When point source of a light is focused to a convex or concave mirror emergent rays make a parallel beam of light.

Answer

(b) $-20cm$

Explanation

Here, size of object $=O=+10.0mm=+1.0cm$ (as, $1cm=10mm)$

Size of Image size $=I=5.0mm=0.5cm$

Image distance, $v=-30cm$ (as image is real)

Let, object distance $=u$

Focal length, $f=$ ?

Magnification $m={\displaystyle \frac{I\text{ (Size of image) }}{\text{ O(Size of image) }}}$

Magnification is given by $m={\displaystyle \frac{-v}{u}}$

${\displaystyle \frac{I}{o}}={\displaystyle \frac{-v}{u}}$

${\displaystyle \frac{0.5}{1}}={\displaystyle \frac{-30}{u}}$

$U=-60cm$

Focal length is given by ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}$

$\begin{array}{rl}{\displaystyle \frac{1}{f}}& ={\displaystyle \frac{1}{-30}}+{\displaystyle \frac{1}{-60}}\\ & ={\displaystyle \frac{-2-1}{60}}\\ & ={\displaystyle \frac{-3}{60}}\end{array}$

$F=-20cm$

Answer

(c) When object is placed between the focus and centre of curvature

Explanation

When object is placed between $F$ and $C$ an enlarged image is formed beyond $C$.

Answer

(a) $3/2$

Refractive Index of B with respect to A

$={\displaystyle \frac{\sin i}{\sin r}}$

$={\displaystyle \frac{\sin 60}{\sin 45}}$

$={\displaystyle \frac{(\sqrt{3}/2)}{(1/\sqrt{}2)}}$

$=\sqrt{6}/2$

$=1.225$

Answer

(b) less than unity

Explanation

Here the ray of light bends away from normal when it enters from medium $A$ into medium $B$. This shows that medium $B$ is optically rarer than medium $A$. Therefore, speed of light in medium $B$ is more than speed of light in medium A. So, ratio of speed of light in medium A to speed of light in medium B will be less than one.

Answer

(a) A rectangular glass slab

Explanation

When incident rays fall perpendicularly on the point of incidence. A rectangular glass slab would refract and then re-refract it.

Answer

(a) Concave lens

Explanation

concave lens because incident rays are parallel and emergent rays are diverging.

Answer

(a) A convex lens has 4 dioptre power having a focal length $0.25m$

Explanation

Positive value for focal length indicates convex lens.

Answer

(a) is less than one

Explanation

Convex mirror is used in rear view mirror. Convex mirror always give smaller image. Hence, magnification produced by rear view mirror is always less than 1 .

Answer

(d) more than $30cm$ in front of the mirror

Explanation

Here $f=15$,

Radius of curvature is double the focal length

hence $c=30$

When object is placed on c, its image is of the same size, inverted and is formed on c.

Answer

(b) a convex mirror

Explanation

Field of convex mirror is more than any type of mirror. Hence full length size of building can be seen by using a convex mirror.

Answer

(b) very near to the focus of the reflector

Explanation

Headlight reflectors and search lights are in the shape of concave mirror. When source of light is placed at the focus, reflected light appears like a beam.

Answer

(d) all mirrors irrespective of their shape

Answer

b)B

Explanation

Light bends towards normal when it passes from air to glass. Light bends away from normal when it passes from glass to air. This is appropriately shown in figure b).

Answer

(d) Glycerine

Explanation

Refractive indices

Water-1.33

Kerosene-1.44

Mustard oil-1.46

Glycerine-1.47

Hence Glycerine is optically dense hence ray of light bends more with glycerine.

Answer

(d) Fig. D

Explanation

In case of concave mirror, an incident ray is parallel to principal axis passes through $F$ after reflection.

Answer

(a) Fig. A.

Explanation

In convex lens, incident ray passing through $F$ goes parallel to the principal axis after refraction.

Answer

(c) Concave, plane and convex

Explanation

When the object is between F and P of concave mirror enlarged image is formed behind the mirror. Hence child can see her head bigger in a concave mirror. She can see her body size of the same size because plane mirror gives image of original size. Convex mirror gives diminished images and babies legs appear smaller.

Answer

(d) Concave mirror, convex mirror, concave lens and convex lens

Answer

a) Concave mirror

b) Convex lens

c) Concave lens

d) Convex mirror

Answer

When light ray enters denser medium from rarer medium it bends towards the normal In this case extent of bending of ray at opposite parallel is same. Hence emergent ray is parallel to incident ray.

Answer

Bending of light here is a function of refraction. Refraction is dependent on refractive indices. Refractive indices of kerosene or turpentine would not be same as water. Hence degree of bend would be different in different mediums.

Answer

Refractive Index can be seen as the factor by which the speed and the wavelength of the radiation are reduced with respect to their vacuum values.

$w=civ$ (where n:refractive index,c=speed of light,v:velocity of light in that medium)

Refractive index of one medium in relation to a second medium is given by ratio of speed of light In second medium to speed of light in first medium.

Answer

Absolute RI of diamond $=1.6$

Absolute RI of glass $=1.5$

Multiplying them we get 2.4

Answer

When an object is placed in F and F2 of a convex lens, we get inverted, enlarged and real image is formed beyond 2F2 which is on the other side of the lens. Hence we need to place the object between 20 and $40cm$ of the lens.

When an object is placed between $F$ and 0 of a convex lens, its enlarged, erect and virtual image is formed beyond FL i.e. on the same side of lens. So for this we need to place the object at a distance less than $20cm$ from the lens.

Answer

To obtain clear image of the building Sudha has to move the screen towards the elns. Focal length will be approximately $15cm$. The rays of light coming from distant object such as a tree (or a distant building or electricity pole) can be considered to be parallel to each other. When parallel rays of light are incident on a convex lens, the rays, after refraction, converge at focus on the other side of the lens.

Answer

Power of lens is inversely proportional to the focal length of the lens. Lens with focal length 20 has more power than lens with focal length $40cm$. Lens with higher power should be used to obtain more convergent light.

Answer

If two plane mirrors are placed perpendicular to each other then Incident ray and reflected ray will always be parallel to each other.

Answer

(i) When a ray of light enters from air (rarer medium) into water (denser medium), the ray of light bends towards the normal.

(ii) When a ray of light enters from water (denser medium) into air (rarer medium), the ray of light bends away from the normal.

Note: The dotted line shows the undeviated path of light, in case if the light would not have undergone refraction.

Answer

a)

C)

d)

Answer

a)

c)

d)

Answer

Laws of refraction

a) Incident ray, refracted ray and normal at the point of incidence lie in the same plane.

b) Ratio of sine of incidence and sine of refraction is constant for the given color and pair of media.

• $ABCD$ is a glass slab.

$EF$ is incident ray which is incident on point 0 on air-glass interface.

• NO is normal and _LEON = :I; which Is angle of incidence.
• NV is normal extended towards the glass slab and ..LV ’ $00=$ Zri; which is angle of refraction.
• 00 ’ is refracted ray from surface $AB$. It behaves like Incident ray on surface $CD$.
• the ray EF bends when it enters the slab to become 00'.
• $M{O}^{\prime }$ and $O^{\prime }M$ ’ are normal on surface $CD$.
• GH is the emergent ray.
• $ZO{O}^{\prime }Al=Li$; ; which is angle of Incidence at surface CD.
• Z. 110 ’ $H=Zr$; which Is angle of refraction at surface CD.
• It is observed that EF. NO and ${00}^{\prime }$ lie in the same plane: which is in accordance to the first law of refraction.
• It is also observed that EF II GH; which means emergent ray is parallel to incident ray. This happens because the degree of bend at opposite surfaces of glass slab is same.

Answer

(a) The ray diagram when the object is placed at the focus of the concave lens:

(b) The ray diagram when the object is placed between focus and twice the length of focal length of the lens:

Answer

a) At Infinity

Answer

As the image is obtained on the screen, it is real .

so,

Magnification , $m=-3$,

$v=80cm$

$u=$ ?

As $m=v/u$

so,

$-3=80/u$,

$u=-80/3cm$.

From

$1/f=1/v-1/u$

$=1/80+3/80$

$=4/80$

$=1/20$

$1/f=1/20cm$

so,

$f=20cm$.

The lens is convex and image formed at $80cm$ from the lens is real and inverted.

Answer

$m={\displaystyle \frac{1}{3}}$

Using ${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$

Calculate $u;u=-80cm$.

Image is real and inverted. Mirror is concave.

Answer

Degree of convergence and divergence provided by a lens is called power of the lens. Unit of power of lens is Diopter D.

focal length of lens used by first student is in positive hence it is a convex lens. The lens of second student is a concave lens.

$p=={\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{0.5}}=2$

Power of lens (first student) $=+2$

Power of lens (second student) = ). 2

Answer

Position of the candle flame $=12.0cm$

Position of the lens $=50.0cm$

Position of the screen $=88.0cm$

i) $u=50-12=38cm$.

Image distance $v=88-50=38cm$

Focal length $={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$

$f=19cm$

ii) Object distance $u=50-31=19cm$

Here

Object distance $=$ focal length

Hence the I mage is formed at infinity.

iii)

If he further shifts the candle towards the lens. The object comes between F and 0 . In this case. Image is virtual, enlarged and erect and is formed on the same side of lens. iv)