Class 10 Science Chapter 9 Previous Year Questions - Ray Optics - Light : Reflection and Refraction

Ans: (A) (i) and (ii)

• Brief explanation: Mirror X is a concave mirror used to concentrate sunlight — for parallel rays from the Sun it forms a real, highly diminished image at the principal focus. Mirror Y is a convex (rear-view) mirror — it always gives a virtual, diminished, erect image.

Ans: 

Use lens formula

For a concave lens take u =−60 cm (object on left) and f = −30 cm. Substitute:

so v = -20 cm

The image is formed at 20 cm on the same side as the object (i.e. v = -20 cm.

Nature: the image is virtual, erect and diminished.

Ans:

(i) Object at infinity:

For a convex mirror, when the object is at infinity, incident rays are parallel to the principal axis. After reflection, these rays diverge and appear to come from the focus behind the mirror.

• Position: At F (behind the mirror)

• Nature: Virtual and erect

• Size: Highly diminished (point-like)

(ii) Object between infinity and pole P:

• For any object in front of a convex mirror, the image is always formed between the pole and focus.

• Position: Between P and F (behind the mirror)

• Nature: Virtual and erect

• Size: Diminished

1. The nature of the image formed:

The negative value of v (i.e. v = −40 cm) indicates that the image is formed on the side opposite to the incident light direction. In this convention, that means the image is real.

The magnification m = −2 is negative, which tells us that the image is inverted relative to the object.

2. Size of the image compared to the size of the object:

The magnitude of the magnification is |m| = 2.

Therefore, the image is twice as large as the object.

(a) (iii):
Focal lengths: f_A = 10 cm = 0.10 m, f_B = 20 cm = 0.20 m.
Power P = 1/f
 so

The chapter states a lens with shorter focal length bends rays through larger angles (greater convergence/divergence) and power is the reciprocal of focal length.
So lens A (10 cm) shows the higher degree of convergence/divergence because it has the larger power (10 D vs 5 D).

OR

(b)(i) 

(b) (ii) Snell’s law of refraction
(i) The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

(ii) for a given pair of media and light colour. This constant is the refractive index of the second medium with respect to the first (Snell’s law).

(b)(iii) Difference between virtual images of a convex lens and a concave lens

Ans: The image is formed at 60 cm in front of the mirror
Given:

• Object distance, u = -30 cm (negative due to the new Cartesian sign convention).
• Focal length of concave mirror, f = -20 cm (negative for concave mirror).
• Mirror formula:

Rearranging for image distance v:
1/v = 1/f - 1/u
Substitute the values:
1/v = 1/(-20) - 1/(-30) = -1/20 + 1/30 = -3/60 + 2/60 = -1/60
v = -60 cm

The negative sign indicates the image is formed in front of the mirror (real image). Thus, the image is at 60 cm in front of the mirror.

Ans: Use the mirror formula
For a concave mirror take u = -10 cm (object to the left) and f = -15 cm. Substitute:
so, v = +30 cm.
The image is formed 30 cm behind the mirror. (i.e v = +30 cm.)
Nature: since the object lies between P and F, the image is virtual, erect and enlarged.

Ans: (a) 10/9
Refractive index of Y w.r.t X
 

Ans: (b) -20 cm

Using

So f = -20 cm. 

Ans:

Given: h = 2.0 cm, h' = 8.0 cm, u = -6.0 cm (object distance taken negative by the New Cartesian sign convention).

Step 1: Magnification:

- For a convex lens, object distance is usually negative if it is to the left (real object). Here, image is on the same side as object. Since both object and image are on the same side, the image is virtual.
Step 3: Using formula for magnification: 

v = −mu

Substitute:

v = −4 × (−6.0) = 24.0 cm

So, the image distance is +24 cm, meaning image is on the opposite side of the lens from the object. This conflicts with the problem statement that both are on the same side.

Since the question says both are on the same side of the lens, and magnification is positive (upright), then for convex lens object is at -6 cm, image should be on the same side (virtual image) with positive magnification.

If we take object distance u=+6 cm (because of sign conventions sometimes vary for virtual object), thenand
Image distance v =−24 cm indicates image is on the same side as object (virtual image).
Step 4: 

 cm:

Ans: (a)

Ans: Given: convex mirror, $\hspace{0.17em}mf=+1.5\backslash ,\backslash text\{m\}$f = +1.5m; object distance $\hspace{0.17em}mu=-6.0\backslash ,\backslash text\{m\}$u = −6.0m; object height $\hspace{0.17em}mh\_o=3\backslash ,\backslash text\{m\}$h_o = 3m.

Mirror formula

v>0 ⇒ image is behind the mirror (virtual).

Magnification

Ans: (c) 40 cm

Image on a screen ⇒ real image (thus a convex lens).
For a convex lens, magnification $m=\frac{v}{u}=-1m=\backslash dfrac\{v\}\{u\}=-1$ ⇒ $v=-u$v = −u.
Using Lens Formula:

Solve for u: (Sign ‘−’ means the object is placed on the incoming-light side; distance = 40 cm.)

⇒ object is 40 cm from the lens.

Ans: (c) Glass slab

When a parallel beam of light passes obliquely through a rectangular glass slab, the emergent ray is parallel to the incident ray but laterally displaced — a characteristic property of a glass slab.

Ans: Use the mirror formula and magnification relation.
Given f = –18 cm (concave), required m = +2.
For mirrors Mirror formula:

Substitute v = -2u

Hence f = 2u so u = f/2  = -18/2 = -9 cm.
the object must be placed 9 cm in front of the mirror (i.e u = -9cm) which is between the pole and the principal focus.
The image will be virtual, erect and magnified (since m = +2); its position is v = -2u = +18 cm (i.e. 18 cm behind the mirror).

Ans: (b) between pole and focus of the mirror

Ans:
(a) Optical density: The optical density of a medium is directly related to its absolute refractive index. Since kerosene has a higher refractive index (1.44) compared to water (1.33), kerosene has a higher optical density.
(b) Mass density: Optical (refractive) index does not determine mass density. In fact, water is more mass-dense than kerosene (kerosene floats on water).
→ No direct relation between optical density and mass density.
(c) Relative speed of light: Speed ∝ 1 / refractive index, so light travels slower in kerosene than in water.
Numerical ratio:

Hence light in kerosene is about 0.924 times as fast as in water.

Ans: (c)

To get an image of the same size as the object using a convex lens, the object should be placed at twice the focal length of the lens. This distance is called "twice the focal length" because at this position, the light rays coming from the object will converge to form an image that is equal in size.

Ans:

Given:

• Object distance, $u=-10\hspace{0.17em}cm\; u\; =\; -10\; \backslash ,\; \backslash text\{cm\}$ (negative, object in front of mirror, Cartesian sign convention).
• Focal length, $f=+15\hspace{0.17em}cm\; f\; =\; +15\; \backslash ,\; \backslash text\{cm\}$ (positive for convex mirror).

Using the mirror formula:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\backslash frac\{1\}\{f\}\; =\; \backslash frac\{1\}\{v\}\; +\; \backslash frac\{1\}\{u\}$
Substitute:
$\frac{1}{15}=\frac{1}{v}+\frac{1}{-10}\backslash frac\{1\}\{15\}\; =\; \backslash frac\{1\}\{v\}\; +\; \backslash frac\{1\}\{-10\}$
$\frac{\mathrm{<br>1}}{v}=\frac{1}{15}+\frac{1}{10}=\frac{\mathrm{2+3}}{30}=\frac{5}{30}=\frac{1}{6}\backslash frac\{1\}\{v\}\; =\; \backslash frac\{1\}\{15\}\; +\; \backslash frac\{1\}\{10\}\; =\; \backslash frac\{2\; +\; 3\}\{30\}\; =\; \backslash frac\{5\}\{30\}\; =\; \backslash frac\{1\}\{6\}$
$v=+6\hspace{0.17em}cmv\; =\; +6\; \backslash ,\; \backslash text\{cm\}$

The image is formed 6 cm behind the convex mirror (positive $vv$ indicates a virtual image).

Ans: 

i.) Cases:

• Mirror A: $f=20,u=\mathrm{45\; cm }$
• Mirror B: $f=15,u=\mathrm{30\; cm}$
• Mirror C: $f=30,u=\mathrm{20\; cm }$

For a concave mirror, a diminished image is formed when the object is beyond the centre of curvature ($u>2f$) or at the centre of curvature ($u=2f$).

• Mirror A: Centre of curvature $2f=40$cm. Since $u=\mathrm{45\; cm}>40$ cm, the image is real, inverted, and diminished.
• Mirror B: Centre of curvature $2f=30$ cm. Since $u=30=2f$, the image is real, inverted, and same-sized (considered diminished as it is not magnified).
• Mirror C: $u=20<f=30$ cm. The object is between the pole and focus, producing a virtual, erect, and magnified image.

Thus, Mirrors A and B form diminished images.

ii.) For Mirror B: $f=15\hspace{0.17em}cm,u=\mathrm{30 }$cm.

• The object is at the centre of curvature ($u=2f=30$ cm).
• Properties:
  1. Real and inverted: The image forms on the same side as the object and is inverted.
  2. Same size: The image is the same size as the object, located at the centre of curvature.

OR
(iii) (B) Here ƒ = –12 cm, u = –18 cm, n = ?

Mirror formula:

v = -36 cm

Ans: (a) Laws of Refraction of light :
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media.
OR
(b) Absolute refractive index of a medium is the ratio of speed of light in air/vacuum to the speed of light in the given medium.
Given:

c = 3 × 10⁸ m/s; n_m = 1.5; v_m = ?

Absolute refractive index of a medium (n_m):

n_m = speed of light in vacuum (c)speed of light in medium (v_m)

n_m = cv_m

v_m = cn_m = 2 × 10⁸ m/s

Ans: (b)

Refraction:

Light bends when entering a different medium, like when sunlight enters a raindrop. 

Dispersion:

Different colors of light bend at slightly different angles within the raindrop, causing them to separate. 

Total internal reflection:

Light is trapped inside the raindrop and reflected back out if the angle of incidence is greater than the critical angle. 

Ans: (a)

Ans: (a) (i) S. No. 3, 2f is 50 cm. ∴ 2f = 50 cm, or  f = 25 cm.
Justification: Object distance(u) and image distance (v) are same so it implies that object is placed at 2F.
(ii) S. No. 6, is incorrect.
Reason: For u = −15 cm, sign of v must be – ve ( as the image is formed on the same side of the lens as the object)

(iii) Magnification: m = v/u
= +150-30 = - 5 .
OR
(b) (i) Principal axis: It is an imaginary line passing through the two centres of curvatures of a lens.  

(ii) f = − 20 cm; h = 5 cm; v = −15 cm

1u = 1v - 1f

or

1u = 1-15 - 1-20

1u = -115 + 120

1u = -460 + 360

1u = -160

u = -60 cm

or u = − 60 cm object is at a distance of 60 cm from the lens
Size of the image(magnification):m = h'h = vu
h' = vu × h = -15-60 × 5 = 1.25 cm

Ans: (c)
If the upper half of a convex lens is wrapped in black paper, it blocks some light from passing through. As a result, the brightness of the image formed will reduce, but the size and shape of the image will remain the same since the lower half of the lens can still focus the light.

Ans: (a) The Principal axis of a concave mirror is an imaginary line that runs through the centre of the mirror, perpendicular to its surface. It is the line along which light rays parallel to it converge after reflection.
(b) Relation between the focal length (f) and the radius of curvature (R):$R=2fR\; =\; 2f$

Given F = 10 cm, therefore R = 20 cm.

Radius of curvature ,R= 20 cm
(c) (i) u = -10 cm, f = +15 cm

1f = 1v + 1u

1v = 1f - 1u = 115 - 1-10

1v = 115 + 110

1v = 16

∴ v = +6 cm

The image is formed at a distance of +6 cm behind the mirror.

Magnification:

Image height:

Final Answer:

• Position of image: $-48\hspace{0.17em}cm\backslash boxed\{-48\; \backslash ,\; \backslash text\{cm\}\}$−48cm (on same side as object → virtual and erect)

• Size of image: $+12\hspace{0.17em}cm\backslash boxed\{+12\; \backslash ,\; \backslash text\{cm\}\}$+12cm (3 times taller than the object)

Ans: (a)

The refractive index of glass is lowest for red light, meaning that red light travels through glass the fastest compared to other colors. As a result, red light bends the least when it enters the glass, which is why it has the minimum refractive index.

• Object distance: $u=-20$cm (negative, as the object is in front of the mirror).
• Magnification: $m=+1/2$ (positive, suggesting a virtual, erect image, typical for a convex mirror).
• Magnification formula:The image is at $$v=+10cm, virtual and behind the mirror.
  Find Focal Length:
  • Use the mirror formula
    The positive focal length indicates a convex mirror, consistent with the positive magnification.
    • New Magnification Requirement:
      • New magnification: $m\prime =+1/3$
      • Magnification formula:

        Mirror Formula: 

        

        The new object distance is 40 cm from the mirror.

    • Verification:
      Calculate $v\prime <br><img\; src="https://cn.edurev.in/ApplicationImages/Temp/26301969\_6500539a-6f24-40ba-9379-b3ccfc68ee88\_lg.png"\; style="max-width:\; 100\%;\; word-break:\; break-word;\; width:\; 379px;\; cursor:\; pointer;\; padding:\; 0px\; 1px;\; user-select:\; none;\; display:\; block;\; margin:\; 5px\; auto;\; text-align:\; center;\; letter-spacing:\; inherit\; !important;"\; alt="Class\; 10\; Science\; Chapter\; 9\; Previous\; Year\; Questions\; -\; Ray\; Optics\; -\; Light\; :\; Reflection\; and\; Refraction">$

      Magnification:
      The magnification is +1/3, as required.
      The object should be placed at a distance of 40 cm from the spherical mirror to reduce the magnification to $+1/3+1/3$+1/3.

Ans:  For Diverging mirror (Convex Mirror)
(i) Principal focus: It is the point on the principal axis where rays incident parallel to the principal axis appear to diverge after reflection.
(ii) Focal length: The distance between the pole of the mirror and the principal focus is called focal length.

Ans: 

• Image distance: $v=+37.5\hspace{0.17em}cm\; v\; =\; +37.5\; \backslash ,\; \backslash text\{cm\}$v=+37.5cm
• Image height: $h\prime =-15\hspace{0.17em}cm\; h\text{'}\; =\; -15\; \backslash ,\; \backslash text\{cm\}$h′=−15cm

Ans: Power:  The ability of a lens to converge or diverge the rays of light is called power of lens. It is the reciprocal of focal length of lens.
P = 1f (cm) = 100f (cm)
Its Sl unit is dioptre.
If f = - 10 cm
P = 100f = -10010 = -10D
So, lens is concave in nature . u = - 20 cm, f = - 10 cm .
As the object placed beyond F image is virtual and thus magnification is +ve.

Ans: (i) Refractive index of diamond,

n = Speed of light in vacuumSpeed of light in diamond

n = cv

2.42 = 3 × 10⁸Speed of light in diamond

Speed of light in diamond = 3 × 10⁸2.42

= 1.25 × 10⁸ m/s

(ii) r_water < r_glass < r_{carbon disulphide} 
(iii) (A) (a) Since the speed of light is greater in water than in glass, glass is optically denser than water. This demonstrates that glass presents a greater barrier to light transmission than water.
(b) When a ray of light is incident normally at the water-glass interface, it passes straight through without deviation. This happens because the angle of incidence is $0\circ 0^\backslash circ$0∘, and according to Snell's law, there is no refraction at normal incidence. Only the speed of light decreases in the glass due to its higher refractive index.
OR
(B) Refractive index of glass, η_g = 4/3

Speed of light in vacuum
Refractive index of water

Speed of light in water = 1.73 x 10⁸ m/s
Because the information provided is wrong, ideally the speed of light in vacuum is 3 × 10⁸ m/s and the speed of light in water is 2.25 × 10⁸ m/s.
The correct solution is

Refractive index of glass, η_g = 32

Refractive index of water, η_w = 43

Refractive index of glass, η_g = Speed of light in vacuumSpeed of light in glass

32 = Speed of light in vacuum2 × 10⁸

Speed of light in vacuum = 3 × 2 × 10⁸2 = 3 × 10⁸ m/s

Refractive index of water, η_w = Speed of light in vacuumSpeed of light in water

43 = 3 × 10⁸Speed of light in water

Speed of light in water = 3 × 3 × 10⁸4

Speed of light in water = 2.25 x 10⁸ m/s

Ans: (a) Given: P₁ = 4 D, P₂ = -2 D
P = P₁ + P₂ = 4D - 2D
So the lens is convergent in nature.
(b) P = -2.5 D

P = 100f (cm) ⇒ −2.5 = 100f (cm) ⇒ f = −40 cm

(c) P = 0.1 D, u = −20 cm

P = 100f (cm) ⇒ 0.1 = 100f ⇒ f = 1000 cm

OR
(c) Virtual images formed by convex and concave lenses differ in several ways:

• Convex Lens: Forms a virtual image when the object is within the focal length. The image is erect, magnified, and on the same side as the object.

• Concave Lens: Always forms a virtual image. The image is erect, diminished, and on the same side as the object.

Ans: (A) Applications of concave mirrors: 
(1) Concave mirror is used as a shaving mirror when the face is placed close to it so that it is within its focus and we get an erect and magnified image of the face.
(2) Doctors use concave mirror as a headmirror to concentrate parallel rays of light on its focus which enables them to examine body parts such as eye, throat, etc.
(B) Given, f = 15 cm
We know for a mirror,
R = 2f
R = 2 × 15 cm
R = 30 cm
(C)

(D) (i) We can use the mirror formula to find the focal length of the mirror:

1f = 1v + 1u

where, f is the focal length of the mirror, v is the image distance and u is the object distance. Since the image is formed at the same point as the object, v = u = –100 cm (Distances to the left of the mirror are negative). 
Substituting the values, we get:

1f = 1-100 + 1-100

1f = -2100

f = -50 cm

So the focal length of the mirror is –50 cm. (Negative sign indicates that it is a concave mirror). 
(ii) The magnification of the image is: 

m = - vu

where, m is the magnification of the image.
Substituting the values, we get:

m = - -100-100

m = -1

So the magnification of the image is 1. (Negative sign indicates that the image is real and inverted).

Ans: (A)
(B) The image of object obtained in the convex mirror is erect and diminished. This is because a convex mirror always forms a virtual, erect and diminished image of an object. 
(C) The image distance is positive. This is because, the image is formed is behind the mirror

Ans: (b)
Sol: The image formed by a convex mirror is always erect and of smaller in size than object.

Ans: (d)
Sol: It is valid for both concave mirrors and convex mirrors.

Ans: (a)
Sol: Concave mirrors are the mirrors best suited ir solar cookers because concave mirrors are convergent mirror and they are reflect sunlight towards a single focal point.

Ans: (b)
Sol: Magnification : Sign-positive, value-more the 1 because the object is placed between the focus and the pole. So, magnified image will be formed on other side of mirror. Hence, magnification of image formed will have i positive sign and value more than one.

Ans: (d)
Radius of curvature of a converging mirror,  R = 30 cm
Focal Length, f = 30/2 cm = 15 cm
Thus, virtual image can be obtained from the mirror if an | object is placed between pole and focus, i.e., between 0 i cm and 15 cm.

Ans: (b)
Sol: Image formed is enlarged is not true. When  object is placed at C, image formed is real, inverted and of same size as object.

Ans: (c)
Sol: Given, height of object (h) = +4 cm
Object distance (u) = -30 cm (object placed left side of the mirror)
Focal length, f = +10 cm
Mirror Formula,
1/f = 1/v + 1/u or v = uf/u - f
v = −30 × 10−30 × 10 or v = 304 = 7.5 cm

Now, Magnification (m) = −vu = h′h ; m = - −152 x 130 x = h′4

or h' = 1 cm
Hence, height of the image formed is 1 cm.

Ans: (a)
Sol: As the focal length of a concave mirror and a concave lens is taken as negative, both are concave in nature.

Ans: (c)
Sol: If the rays are travelling from far off distance and focused on opposite side of the lens, this is only possible in convex lens.

Ans: (d)
Sol: Here, medium A and C is same and 8 is glass. n₁ = n₃ < n₂

Ans: (c)
Sol: The image of the candle flame is three times larger than the flame itself, which means the magnification is -3 according to the new Cartesian sign convention, where a negative sign indicates that the image is inverted. Since this type of magnification occurs with a convex lens, the correct answer is that the lens is convex and the magnification is -3.

Ans: (c)
Sol: Given, combined power = +1 D
Focal length of one lens, f₁ = 20 cm
Combined focal length, f_combined = 1 m = 100 cm

1f = 1f₁ + 1f₂ ; 1100 = 120 + 1f₂

Solving this equation, we get f₂ = -25 cm Focal length of other lens = -25 cm

Ans: (b)
Sol: As the ray after passing XX 'is diverging therefore, XX' is concave lens and image is formed between O and Y.

Ans: (c)
Sol: Given, object height = 3.0 cm
Let object distance is u and image distance is v.
Case-1: u = -37.5 cm and v = 25 cm
h = -2 cm (real and inverted)
From the lens formula, 1f = 1v - 1u = 125 - 1-37.5
f = +15 cm ... (i)

Case2: u = -25 cm, f = +15 cm (from (i))
By lens formula:

1f = 1v - 1u

115 = 1v - 125

v = +37.5 cm

Now, h₁ = 3 cm and h = ?

Magnification:

m = vu = hh₁

∴ +37.5-25 = h+3

∴ h = -4.5 cm

Ans: (b)
Sol: The image formed by of a concave lens is always virtual, diminished and erect.

Ans: (b)
Sol: As refractive index of a medium increases, more bending of light takes place. For A - B, ratio of refractive indices of A and B is least, so least bending of light takes place for this pair of media.

Ans: (c)
Sol: In A, B and C transparent media, ∠1 > ∠2 as light bends towards the normal and ∠3 < ∠4 as light bends away from the normal. ∠3 = ∠2 , because of alternate angles between two normals.

Ans: (d)
Sol: 

In medium 2, the light ray bends towards the normal, so n₂ > n₁.
Similarly, in medium 3, the light ray bends more towards the normal which indicate that refractive index of medium 3 is greater than medium 2. So, n₃ > n₁

Ans: (d)
Sol: Given, refractive index of medium A, μ_A = 1.5
Refractive index of medium B, μ_B = 1.33
Speed of light in air, c = 3 x 10⁸ m/s
μ = Speed of light in vacuum (c)Speed of light in medium (v)

For medium A:

μ_A = cv_A ⇒ 1.5 = 3 × 10⁸v_A m/s

v_A = 2 × 10⁸ m/s

For medium B:

μ_B = cv_B ⇒ 1.33 = 3 × 10⁸v_B m/s

V_B = 2.25 x 10⁸ m/s
Hence, speed of light in medium A is 2 x 10⁸ m/s and in medium B is 2.25 x 10⁸ m/s.

Ans: (c)
Sol: To get real & magnified image, object should be kept either between 2F and F or at F. If it is kept at F, image will form at infinity which cannot be taken on screen, so object should be kept between 2F and F.

Ans: (b)
Sol: As per the arrangement, an optical device to perform an experiment of light is likely to be a concave lens.

Ans: (c)
Sol: Converging point, f = 20 cm
Power P = ?
As we know that,

Power of lens = 1focal length (in m) = 1f (in m)

P = 10020 ⇒ P = +5 D

Hence, power of convex lens is +5 D.

Ans: (d)
Sol: –40 cm
Reason — Given, convex lens as converging lens
f = 30 cm
From formula,
m = v/u
-3 = v/u [-ve sign as real and inverted image is formed.]
v = -3u
Using formula,
1/f = 1/v - 1/u
We get,

130 = 1-3u - 1u

130 = -1 - 33u = -43u

3u = -4 × 30

u = -1203 = -40 cm

Hence, distance of the object from the lens = -40 cm.

Ans: It gives us the idea about the speed of light in the air and in the glass. It means that speed of light is 1.5 time more in air than the speed of light in the glass. 

Ans: (b)
The bending of light depends on the difference in refractive indices between two media. The smaller the difference in refractive indices, the less the light will bend when passing from one medium to another.
For each pair of media:

• Air-A: The difference in refractive indices is |1.00 - 1.50| = 0.50.
• A-B: The difference in refractive indices is |1.50 - 1.33| = 0.17.
• B-C: The difference in refractive indices is |1.33 - 2.42| = 1.09.
• C-Air: The difference in refractive indices is |2.42 - 1.00| = 1.42.

The smallest difference is between A and B, which has a refractive index difference of 0.17. Therefore, the bending of light will be least when it passes between A and B.
Thus, the correct answer is (b) A-B.

Ans: (d)
Given:
The lens forms a real, three times magnified image (m = -3 because the image is real and inverted).
Focal length $\hspace{0.17em}cmf\; =\; 30\; \backslash ,\; \backslash text\{cm\}$f = 30cm.

For a lens, the magnification $m$m is given by: m = vu

where $\mathrm{<br>}v$v is the image distance and $\mathrm{<br>}u$u is the object distance. Since m = −3:

vu = -3 ⇒ v = -3u

Using the lens formula:

1f = 1v - 1u

Substitute $\hspace{0.17em}cmf\; =\; 30\; \backslash ,\; \backslash text\{cm\}$f = 30cm and v = −3u:

130 = 1-3u - 1u

Simplify this equation:

130 = -1 - 33u = -43u

Now, solve for u:

u = -4 × 303 = -40 cm

Therefore, the distance of the object from the lens is –40 cm.

Ans: (c)
Given:
Height of the object, h = 4cm
Object distance from the mirror, $\hspace{0.17em}cmu\; =\; -30\; \backslash ,\; \backslash text\{cm\}$u = −30cm (distance is negative for mirrors according to the sign convention)
Focal length of the diverging (concave) mirror, $\hspace{0.17em}cmf\; =\; -10\; \backslash ,\; \backslash text\{cm\}$f = −10cm

We need to find the height of the image h′.

1f = 1v + 1u

Substitute the given values into the formula:

1-10 = 1v + 1-30

Simplify the equation:

1v = -330 + 130 = -230 = -115

Therefore:

v = -15 cm

Calculating the magnification (m):

m = h_ih_o = vu

Substitute the known values:

m = -15-30 = 12

Also, To find the height the value of h_i

Using the magnification formula:

m = h_ih_o

Substitute the values of m and h_o:

12 = h_i4

Solve for h_i:

h_i = 12 × 4 = 2 cm

So, the height of the image formed is 2 cm.

Ans: The pole of a spherical mirror define the geometrical center of the spherical surface of the mirror. It is the center of reflecting surface of spherical mirror and lies on the surface of spherical mirror.

Ans: Given, refractive index of medium x with respect to y, vμ_x = 23

Refractive index of medium y with respect to z, zμ_y = 43

∴ Refractive index of medium x with respect to z, zμ_x = v_{μx x} zμ_y = 23 × 43 = 89

∴ Refractive index of medium z with respect to x, xμ_z = 1zμ_z = 98

Now, speed of light in x = 3 × 10⁸ m/s

Speed of light in y, v_y = ?

yμ_x = Speed of light in xSpeed of light in y
⇒ v_y = 23 × 3 × 10⁸ = 2 × 10⁸ m/s

Ans:  (a) We have used a converging lens.
(b) The characteristics of the image formed:
(i) It is real.
(ii) It is inverted
(iii) It is enlarged.
(c) We get the magnification of object, m = - 1 at the  position 2F₁.

Ans: (i) The lens used here is a convex lens and it is used as a magnifying glass because at close range, i.e., when the object is placed between optical centre and principal focus it forms an enlarged, virtual and erect image of the object.
(ii) When this lens is placed such that the object is between the centre of curvature and the principal focus, the palmist obtain a real and magnified image.
(iii) Given, focal length, f = 10 cm and u = -5 cm 
According to lens formula,

1f = 1v − 1u or 1f = 1u + 1v

= 110 + 1−5 = −5 + 10−50

∴ v = −505 = −10 cm

Thus, the image will be formed at 10 cm on the same side of the palm and the size of the image will be enlarged.

Ans: (A) An incident ray parallel to the principal axis, after reflection appear to diverge from the principal focus, in the case of a convex mirror

A straight line XP is principal axis. AB is incident ray which is parallel to principal axis XP of a convex mirror MM’. The ray of light gets reflected at point B on the mirror and goes in the direction BD and it appears to be coming from the focus F of convex mirror. According to the laws of reflection, ∠i = ∠r. Therefore, the incident and reflected rays make equal angles with the normal. 
(B) An incident ray passing through the principal focus of a concave mirror, after reflection, will emerge parallel to the principal axis

(C) An incident ray coming obliquely to the principal axis towards a point (Pole of the mirror) on the convex mirror is reflected obliquely.

Ans: (A)

P = −2.5 D

f = ?

P = 1f

f = 1P = 1−2.5 D

f = −0.4 m = −40 cm

f = −40 cm

v = −10 cm

u = ?

1f = 1v − 1u

1u = 1v − 1f

1u = 1−10 − 1−40

= −110 + 140 = −4 + 140

= −340

u = −403 cm
= -13.33 cm

The object should be placed – 13.33 cm from the lens for the formation of an image at a distance of 10 cm from the lens.
(B)

Ans: (A) When an object is placed in front of the lens between optical centre and principal focus of a convex lens, the image is formed beyond 2F₁ (on the same side of the object).

AB is the object and A’B’ is the image. The image formed is enlarged, virtual and erect. So the value of magnification will be greater than 1 and its sign will be positive. 
(B) When an object a placed anywhere infront of a concave lens. When we place an object between infinity and optical centre (O) of the concave lens, the image will be formed between focus (F1) and optical centre (O) on the same side of the lens.

AB is the object and A’B’ is the image. The image formed is diminished, virtual and erect. So the sign of magnification is positive (+) and the magnification will be less than 1.
(C) When the object is placed at 2F₁ of convex  lens, the image is formed at 2F₂ on the other side of the lens.

The image formed is real, inverted and of the same size.

Ans: Focal length of convex lens (f) = 20 cm
Real image formed at a distance, (v) = 30 cm
Height of image (h₁) = 4 cm
Let the object distance be u.

Lens formula,

1v − 1u = 1f ⇒ 130 − 1u = 120

u = −60 cm

From,

m = vu = 30−60 = −12

m = h'h = −12

h' = −12 × 4

h' = −2 cm

Hence, height of image of object is  -2 cm.

Ans: The laws of reflection of light state that:

• The incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in the same plane.
• The angle of incidence (i) is always equal to the angle of reflection (∠r) i.e. ∠i = ∠r
  

Characteristics of the image formed by a plane mirror

• Image formed in a plane mirror is virtual and erect.
• It is the same size as the object.
• The image formed is laterally inverted.
• The image formed is at same distance behind the mirror as object is in front of mirror.

Ans: 
(a) The student should adjust the distance between the mirror and the paper so that solar rays are sharply focussed on the paper.
(b) The mirror is a concave mirror.
(c) The student can find the approximate focal length by measuring the distance between the paper and the mirror.

As shown in Fig. 10.29, parallel rays from the sun are focussed on the paper at point A' in focal plane of mirror such that PB' = f.

Ans: Here distance of the object from the lens u = - 12 cm and magnification of real image m = -2/3
As per relation m = v/u, we have v = mu =(-2/3) x (-12) = + 8 cm

So as per lens formula 1v - 1u = 1f

We have 1f = 1(+8) - 1(-12) = 18 + 112 = 524

Hence, f = 245 = 4.8 cm

Ans: A ray diagram showing refraction through a rectangular glass slab has been shown in adjoining Fig.

The emergent ray GH is exactly parallel to the incident ray EFNM. It means that ∠r₂ = ∠i₁.
However, the emergent ray is laterally (side ways) displaced as compared to the original path of light ray. In ray diagram, the lateral displacement is GN. Its value increases on increasing the width of glass slab.

Ans: We have, (i) Object distance, u= —60 cm Focal length of the concave lens, f= —30 cm Using lens formula,

So as per lens formula 1v - 1u = 1f

We have 1v = 1(-60) = 1(-30)

Next, 1v = -130 - 160

Then, 1v = -360

Thus, v = -20 cm
The image will be formed at a distance of 20 cm in front of the lens.
(ii) Nature of the image is virtual. The position of the image is between F1 and optical center 0. Size of the image is diminished. The image is Erect.
(iii)

Ans: (a) When an object is placed between the optical centre and principal focus of a convex lens, the image formed is virtual, erect and enlarged. Moreover, the image is formed on the same side of lens behind the object.
(b) Here magnification of given concave lens m = 1/3 and focal length of lens f = - 20 cm
As per relation m = v/u for a lens, we get

+13 = vu ⇒ vu = + u3

Therefore, as per sign convention followed, both u and v are -ve.

Using lens formula 1v - 1u = 1f, we have

1( -u/3 ) - 1( -u/20 ) ⇒ -3u + 1u = -120 ⇒ -2u = -120

Therefore, u = 40 cm

So the object is placed at a distance of 40 cm from the lens.

Ans: Laws of refraction: 
a. The incident ray, refracted ray and normal to the point of incidence all lie in the same plane.
b. The ratio of sin of incident angle to sin of angle of refraction for a given pair of medium is constant.
Sin iSin r = Constant
Absolute refractive index of a medium is the ratio of speed of light in air or vacuum to speed of light in the medium.
Absolute refractive index
= Speed of light in air/vacuumSpeed of light in Medium

Ans: Power of lens is the ability of a lens to converge or diverge light rays passing through it. It is the reciprocal of the focal length. S.I. Unit: The Unit of power of a lens is Dioptre (D)

P = 1f(D)

Power of first lens: Focal length = +40 cm. Focal length is positive, hence it is a convex lens.

P₁ = 100f(cm) = 10040 cm = +2.5 D

Power of second lens:

Focal length = -20 cm. Its focal length is negative, hence it is a concave lens.

f₂ = -20100 m = -15 m

P₂ = 1f₂ = -5D

Ans: Ray diagram: Position of O and F

1f = 1v - 1u ⇒ 120 = 1v - 1(-30)

1v = 120 - 130 ⇒ 1v = 360 - 260

Thus, v = 60 cm

m = h_ih_o = vu ⇒ h_i4 = 60(-30)

Ans: Only a convex mirror always form an erect and diminished image behind the mirror between its pole and focus point for all positions of the object placed in front of the mirror.

A ray diagram showing the image formation of an object AB is shown here . A convex mirror is used as a rear view mirror in automobiles because it gives erect and diminished images of vehicles coming from behind. As a result, it helps the driver in having a much wider field of view.

Ans: Here distance of object u = - 15 cm, height of object h = +4 cm and the focal length of concave mirror f = - 10 cm.

As per mirror formula 1v + 1u = 1f

We have 1v = 1(-10) = 1(-15) = 130 ⇒ u = -30 cm.

Thus, a screen is placed in front of the mirror at a distance of 30 cm from it.

Hence, magnification m = h_ih_o = -vu

Thus, h = -vu × h = (-30)(-15) × 4 = -8 cm.

Thus, the image is an inverted image of height 8 cm.

Ans: Here, object distance, u = -15 cm
Using lens formula,

1f = 1v - 1u

1-30 = 1v - 1-15

1v = 1-30 - 115

Thus, v = -10 cm
Magnification, m = u/v = -10/-15 = +23
Four characteristics of the image formed by the concave lens are :
(i) Image formed is virtual
(ii) Image is erect
(iii) Image is formed on the same side of the lens as the object
(iv) Image is smaller than the object

Ans: (i) Concave mirrors:

• Concave mirrors are used while applying make–up or shaving, as they provide a magnified image.
• They are used in torches, search lights, and head lights as they direct the light to a long distance.

(ii) Convex mirrors:

• Convex mirrors are used in vehicles as rearview mirrors because they give an upright image and provide a wider field of view as they are curved outwards.
• They are found in the hallways of various buildings including hospitals, hotels, schools, and stores. They are usually mounted on a wall or ceiling where hallways make sharp turns.

Ans: Power is the ability of lens to converge or diverge the ray of light is called power of lens. It is equal to the reciprocal of focal length, i.e. P = 1/f

Ans: Magnification, $m=-1m\; =\; -1$
Image distance, $v=\mathrm{25 <br>}$
Calculate Object Position:
For a real image, $m=-1m\; =\; -1$
Magnification formula:

So, the object is placed at $u=-\mathrm{25.<br>}$
v=+25cm indicates a real image (right of lens), $uu$u is negative (object left of lens).Calculate Focal Length
Using the lens formula:
The positive focal length indicates the lens is convex with a focal length of $\mathrm{12.5\; cm.<br>}$New Image Position After Displacement
If the object is shifted $15\hspace{0.17em}cm\; 15\; \backslash ,\; \backslash text\{cm\}$15cm towards the optical centre, the new object distance is:$u\prime =-25+15=-10\hspace{0.17em}cm\mathrm{.<br><img\; src="https://cn.edurev.in/ApplicationImages/Temp/26301969\_b107ef62-7cec-4cdc-a5e2-f88f0c704a16\_lg.png"\; style="display:\; block;\; margin:\; 5px\; auto\; 5px\; 0px;\; text-align:\; left;"\; alt="Class\; 10\; Science\; Chapter\; 9\; Previous\; Year\; Questions\; -\; Ray\; Optics\; -\; Light\; :\; Reflection\; and\; Refraction">}$

Ans: It is a concave or diverging lens.

f = 1/P
P = - 10 D,
f = 1/-10D = -0.1 m
Or f = -10cm.

Ans: Magnification of spherical mirror (m): It is equal to the ratio of size (height) o f the image to the size (height) of the object. Thus,
m = Size of Image (h₂)/Size of Object (h₁)
Given: For concave mirror u = - 20 cm,v = - 40 cm,
Using mirror equation,
1/f = 1/v + 1/u
or
1/f = 1/-40 + 1/-20
= -(1/40 + 1/20)
= -3/40

⇒ f = - 40/3

Ans: Snell’s law: The ratio of sine of angle of incidence (i.e. sin i) to the sine of angle of refraction (i.e. sin r) is always constant for the light of given colour and for the given pair of media.
Mathematically, sin i/ sin r = constant = n₂₁
The constant n₂₁ is called refractive index of the second medium with respect to the first medium.
Absolute refractive index of the medium is given by n_m = Speed of light in vacuum (c)/ Speed of light in medium  (v) = c/v

Ans: (a) Two rays are required.

(b) The linear magnification produced by a spherical mirror is +3. It shows that the size of image is three times the size of object, image is virtual and erect and formed behind the mirror.
Hence
(i) the mirror is a concave mirror, and (ii) the object is placed between the pole and the focus of a concave mirror.(c) The four characteristics of the image formed by the convex mirror are virtual, erect, diminished and laterally inverted.

Ans: (a) Rays which are chosen to construct a ray diagram for reflection are:  A ray parallel to the principal axis and  A ray passing through the centre of curvature of a concave mirror.
Path of these light rays after reflection:
(i) It will pass through the principal focus of a concave mirror
(ii) It gets reflected back along the same path. When an object is placed between the pole and the principal focus of a concave mirror, a virtual, erect and enlarged image is formed behind the concave mirror as shown in the figure.

(b) m=−3, 
$u=-20\hspace{0.17em}cm\; u\; =\; -20\; \backslash ,\; \backslash text\{cm\}$ cm , 
$v=?v\; =\; ?$ 

Image Position

Distance Between Screen and Object

Answer:
Distance = $40\hspace{0.17em}cm\; 40\; \backslash ,\; \backslash text\{cm\}$ cm.

Ans: (a) From the observation 3, the radius of curvature of the lens is 40 cm as distance of object and the distance of the image is same.
We know, focal length, f = R/2 = 40/2 = 20 cm
(b) Observation 6 is not correct, because for this observation the object distance is between focus and pole and for such cases, the image formed is always virtual. But in this case real image is formed as the image distance is positive.

From the figure, object distance u = - 60 cm and image distance v = 30 cm.
We know, magnification = v/u = +30/-60 = -0.5

Ans: (a) Convex (diverging) mirror

Reason: (i) It always produces a virtual and erect image.
(ii) The size of image formed is smaller than the object.

Therefore, it enables the driver to see a wide field view of the traffic behind the vehicle in a small mirror.
(b) Radius of Curvature: The separation between the pole and the centre of curvature or the radius of the hollow sphere, of which the mirror is a part, is called radius of curvature (R), i.e., PC = R.
Since focal length of the mirror is +24 cm. It indicates that nature of the given spherical mirror is convex/diverging mirror.
As R = 2f = 24 cm
Therefore, f = +12 cm

Ans: (a) A convex lens of focal length 'f' can form 
(i) a magnified and erect image only when the object is placed between its focus 'F' and optical centre ‘O’ of the lens.

(ii) a magnified and inverted image when an object is placed in the following positions: Between F₁ and 2F₁

(b) Whatever be the position o f object as given in case (i) and (ii),the image formed by the concave lens is always virtual, erect and diminished.

Ans: Laws of reflection
(i) The incident ray, reflected ray and normal to the reflecting surface at the incident point all lie in the same plane.
(ii) The angle of reflection is equal to the angle of incidence.

Characterstics of the Image:

Ans: (c)
In an experiment involving a rectangular glass slab, when light passes through it, the angle of incidence (∠i) and angle of emergence (∠e) are equal. However, the angle of refraction (∠r) inside the glass slab is different due to the change in the medium's refractive index.

For two different angles of incidence ($\angle i\; =\; 30°$∠i = 30° and $\angle i\; =\; 45°$∠i = 45°):

• When ∠i = 30°, the angle of refraction ∠r will be smaller due to the denser medium, resulting in $\angle r\; \approx \; 20°$∠r ≈ 20°. As the ray emerges from the glass, the angle of emergence ∠e will be equal to the angle of incidence, so ∠e = 30°.
• When $\angle i\; =\; 45°$∠i = 45°, the angle of refraction ∠r will be approximately 28°. Again, the angle of emergence ∠e will match the angle of incidence, so $\angle e\; =\; 45°$∠e = 45°.

Therefore, the correct set of values is (c) ∠r = 20º, ∠e = 30º and ∠r = 28º, ∠e = 45º.

Ans: A prism is an optical device with two triangular bases along with three rectangular lateral surfaces commonly inclined at an angle of 60°. 

Ans: The bouncing back of a ray of light in the same medium after striking on a surface of an object.

Ans: a. Convex lens when object is in between F and C (2F).

b. Concave mirror when object is in between P and F its enlarged, erected and virtual image is formed.

Ans: Given f = + 10 cm, u = ?
For virtual image
m = + 2
As
m = v/u or v/u = 2
v = 2u ...(1)
1/v = 1/u = 1/f ...(2)
Substituting (1) in (2)

12u - 1u = 110

12u = 110

u = -5 cm

For real image, f = 10 cm, m = -2

vu = -2, v = -2u

1v - 1u = 1f

1(-2u) - 1u = 110

-3/2u = 1/10 or u = -15 cm

Ans: a. If the lower half of the lens is covered even then it will form a complete real, inverted image of same size at C₂(2F₂) with reduced intensity of image.

b. There will be no change in the nature and position of the object except in later case the image will be brighter.

Ans: For concave or convex mirrors the relation between u, v and f is given by
mirror formula,

1v - 1u = 1f

m = - 2
u = - 10 cm
m = - v/u = - 2 or v = 2u = -20 cm
v = - 20 cm

Ans: (A) Yes, even when one half of the convex lens is covered with a black paper, the complete image of the object will be formed. When the upper half of the lens is covered: In this situation, rays of light coming from the object will be refracted by the lower half of the lens as shown in fig (a). These rays meet at the other side of the lens to form the image of the given object. When the lower half of the lens is covered: In this situation, rays of light coming from the object will be refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the figure (b). We will get a sharp image but the brightness of the image will be less now. 
Ray diagram:

(B) Convex lens can be used as a magnifying glass.

Ans: (A) A real, inverted and same size image as that of object, formed by the concave mirror, will form an image of magnification –1. It is possible only when the object is placed at C (R = 2f). Hence, for the object distances of 20 cm and 30 cm, concave mirrors ‘A’ and ‘B’ will form the real, inverted and same size images as that of the object. Therefore, the mirrors ‘A’ and ‘B’ will form an image of magnification –1.
(B) Concave mirror of focal length 20 cm will be preferred to be used for shaving purpose or makeup. This is because when we bring our face within its focal length it forms a virtual, erect and enlarged image of our face. 
(C) Ray diagram for image formation by mirror ‘B’ 
(i) For object distance 10 cm.

(ii) For object distance 20 cm.

Ans: Magnification of images formed by plane mirrors is unity because for plane mirrors, the size of the image formed is equal to that of the object.

Ans: Power is the degree of convergence or divergence of light rays achieved by a lens.
It is defined as the reciprocal of its focal length. i.e, P = 1/f

Ans: Dentist uses a concave mirror to see large images of the teeth of patients.

Ans: The shifting of the light ray sideways (though in the direction of original ray) on emergence from a rectangular glass slab is called “lateral displacement”.

Ans: Reciprocal of focal length of a lens, expressed in meter, is called the power of that lens.
Its SI unit is 1 dioptre (1 D), where 1 D = 1 m^-1.

Ans: A convex lens can be used as a magnifying glass so as to form magnified image of a tiny object placed near it.

Ans: A concave (diverging) lens has a negative power.

Ans: 'Virtual image formed by a convex lens is always magnified but that formed by a concave lens is diminished one.

Ans: During its passage from one medium to another a light ray changes its path at the boundary face separating the two media.

Ans: Focal length  f = 1/P = +1/5 m = 0.2m

Ans: If the focal length is measured in metre then the unit of power of a lens is dioptre.

Ans: Only in convex mirror, for all positions of the object placed in front of it is always virtual, erected and diminished. Hence this mirror is convex mirror.


Convex mirrors are used in automobiles as a rear view mirror because of wider field of view and formation of erect image.

Ans: a. Concave mirror

b. Convex mirror

Concave mirror form magnified virtual image of an object.

Ans: The characteristics of an image formed by a plane mirror are:

• The image is virtual.
• The image is erect.
• The image is laterally inverted.
• The size of the image is equal to that of the object.

Ans: Spherical mirror is a part of a sphere. If reflection takes place from inside, it is said to be concave mirror, and if the reflection takes place from outside surface it is a convex mirror.

Ans: (a) The mid point of mirror is known as pole.
(b) The centre of curvature of a spherical mirror is the centre of that sphere of which mirror is a part,
(c) The distance between pole and centre of curvature is called radius of curvature of the mirror.
(d) The straight line joining the pole and centre of curvature is called principal axis.
(e) The point on the principal axis through which parallel rays to the principal axis passes or appear to pass after reflection.
(f) The diameter of the mirror or size of the mirror is called aperture.
(g) The distance between focus and pole of a mirror is the focal length of the mirror.

Ans: 

Ans: Given: u = - 12 cm
f = + 15 cm
h₁ = 4.5 cm
v = ?, m = ?

• Image distance: $v \approx +6.67$cm
• Magnification: $m\approx +0.556m\; \backslash approx\; +0.556$
• As needle moves farther, image remains virtual, erect, approaches 15 cm, and shrinks.

Ans: (b)
When light passes through a rectangular glass slab from air, it bends towards the normal upon entering the glass due to refraction. Inside the glass slab, the ray travels in a straight line parallel to the initial direction but displaced laterally. Upon exiting the glass slab into air, it bends away from the normal and emerges parallel to the original incident ray.

In the diagrams:

• Student B shows the correct path of the light ray with lateral displacement and emergence parallel to the incident ray.
• The other diagrams do not show the correct behavior of refraction through a rectangular glass slab.

Thus, the correct answer is (b) Student B.