Draw ray diagrams showing the image formation by a concave mirror when an object is placed:
(a) Object between pole and focus: The image formed is virtual, erect and enlarged behind the mirror. Rays diverge on reflection but appear to meet behind the mirror.
(b) Object between focus and centre of curvature: The image is real, inverted and enlarged, formed beyond the centre of curvature.
(c) Object at centre of curvature: The image is real, inverted and of the same size, formed at the centre of curvature.
(d) Object beyond centre of curvature: The image is real, inverted and diminished, formed between focus and centre of curvature.
(e) Object at infinity: The image is real, inverted and highly diminished, formed at the focus.
Draw ray diagrams showing the image formation by a convex lens when an object is placed:
(a) Object between optical centre and focus: The image is virtual, erect and enlarged, formed on the same side of the lens as the object.
(b) Object between focus and 2F: The image is real, inverted and enlarged, formed beyond 2F.
(c) Object at 2F: The image is real, inverted and same size, formed at 2F.
(d) Object at infinity: The image is real, inverted and highly diminished, formed at the focus.
(e) Object at focus: The image is formed at infinity, real and inverted.
Write laws of refraction. Explain the same with the help of a ray diagram when a ray of light passes through a rectangular glass slab.
Laws of Refraction:
Explanation with diagram: When a ray of light enters a glass slab from air, it bends towards the normal due to decrease in speed. On emerging from the glass slab into air, it bends away from the normal. However, the emergent ray is parallel to the incident ray because the opposite parallel surfaces cause equal deviation in opposite directions.
Draw ray diagrams showing the image formation by a concave lens when an object is placed:
(a) Object at focus: The image is virtual, erect and diminished, formed between the optical centre and focus on the same side as the object.
(b) Object between F and 2F: The image is virtual, erect and diminished, formed between the optical centre and focus.
(c) Object beyond 2F: The image is virtual, erect and diminished, formed closer to the optical centre.
Draw ray diagrams showing the image formation by a convex mirror when an object is placed:
(a) Object at infinity: The image is virtual, erect and highly diminished, formed at the focus behind the mirror.
(b) Object at finite distance: The image is virtual, erect and diminished, formed behind the mirror between pole and focus.
The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
Given magnification \( m = 3 \) and image distance \( v = 80 \text{ cm} \)
Using \( m = \dfrac{v}{u} \), we get:
\[ 3 = \dfrac{80}{u} \Rightarrow u = \dfrac{80}{3} = -\,26.7 \text{ cm} \]
The negative sign shows that the object is placed on the same side of the lens as the incident light. The image at 80 cm is real and inverted. Hence, the lens is convex.
The size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to one-third of its size. At what distance has the object been placed from the mirror? What is the nature of the image and the mirror?
Given magnification \( m = \dfrac{1}{3} \) and focal length \( f = -20 \text{ cm} \) (concave mirror).
Using \( m = -\dfrac{v}{u} \) and the mirror formula \( \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \), we get:
\[ -\dfrac{1}{3} = -\dfrac{v}{u} \Rightarrow v = \dfrac{u}{3} \]
Substituting in the mirror equation:
\[ \dfrac{3}{u} + \dfrac{1}{u} = -\dfrac{1}{20} \]
\[ \dfrac{4}{u} = -\dfrac{1}{20} \Rightarrow u = -80 \text{ cm} \]
Thus the object is placed 80 cm from the mirror. The image is real and inverted. The mirror is concave.
Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of −50 cm. What is the nature of the lenses and the power used by each of them?
Power of a lens: It is the reciprocal of focal length in metres: \( P = \dfrac{1}{f} \).
Unit: Dioptre (D).
For focal length \( f = +50 \text{ cm} = 0.5 \text{ m} \):
\[ P = \dfrac{1}{0.5} = 2 \text{ D} \]
The lens is convex.
For \( f = -50 \text{ cm} = -0.5 \text{ m} \):
\[ P = \dfrac{1}{-0.5} = -2 \text{ D} \]
The lens is concave.
A student focused the image of a candle flame on a white screen using a convex lens. He noted down the positions of the candle, lens and screen. Using the recorded positions, answer the following:
Given:
Position of candle = 12 cm
Position of lens = 50 cm
Position of screen = 88 cm
(i) Focal length:
Object distance \( u = 12 - 50 = -38 \text{ cm} \)
Image distance \( v = 88 - 50 = 38 \text{ cm} \)
Using lens formula:
\[ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \]
\[ \dfrac{1}{38} - \left( -\dfrac{1}{38} \right ) = \dfrac{2}{38} = \dfrac{1}{19} \]
Thus, \( f = 19 \text{ cm} \).
(ii) If the candle is shifted to 31 cm: The image forms at infinity.
(iii) If the candle is moved still farther: The image becomes virtual and erect.
(iv) Ray diagram: (Diagram shows a convex lens with object between focus and optical centre, rays diverging but appearing to meet behind lens forming virtual image.)