CBSE Pathshala

MIRROR FORMULA The relation between object distance(u), image distance(v) and the focal length (f) of the mirror is known as MIRROR FORMULA. Consider a concave mirror MM' (of small aperture). An object AB of height h is placed on the left of the mirror beyond its center of curvature so that the image formed is real, inverted and diminished in size(height = h'). Here : Object distance = PB = -u Image distance = PB'= -v Focal length = PF = -f Radius of curvature = PC = -R As shown in Figure, Δ ABP ∼ Δ A'B'P. ∴ A'B'/AB = PB'/PB ∴ A'B'/AB = (-v)/(-u) ∴ A'B'/AB = v/u.............(I) Similarly, as shown in the figure, Δ ABC ∼ Δ A'B'C. ∴ A'B'/AB = CB'/CB ∴ A'B'/AB = (PC-PB')/(PB-PC)........[as CB'=(PC-PB') and CB=(PB-PC)] ∴ A'B'/AB = [-R - (-v)]/[-u - (-R)] ∴ A'B'/AB = (-R + v)/(-u + R)........(II) By (I) & (II), v/u = (-R + v)/(-u + R) ∴ -...

Example : 1 Calculate the minimum distance between object and its real image formed by convex lens of focal length F (a) f (b) 2f (c) 3f (d) 4f            Answer: Let the distance between object and real image formed by convex lens is d  and Let object image distance is x  ; then image distance is  d –x By using Lense formula, But, 1/v - 1/u = 1/f So, 1/(d-x) - 1/(-x) = 1/f. {x-(d-x)}/{x(d-x)} =1/f {x-d+x}f =xd – x 2 df = dx – x 2 x 2 – dx + df = 0 Using quadratic equation, For real value, D ³ 0 b 2 + 4ac   ³ 0 (-d) 2 – 4x1xdf ³ 0 d 2 – 4df ³ 0 d 2 ³ 0+4df d ³  4f Therefore, Let the distance between object and real image formed by convex lens is 4f. Example : 2 A plane mirror makes an angle of 30o with horizontal. If a vertical ray strikes the mirror, find the angle between mirror and reflected ray [RPET 1997] (a) 30 o (b) 45 o (c) 60 o (d) 90 o  ...

NCERT Solution, MCQs, Study Notes, Q & A Question: What is the minimum distance between an object and its real image in case of a concave mirror? Ans: The object and the image, both are at the centre of curvature of the concave mirror. The image formed is real and inverted.  The minimum distance  between object and its image =  0 units Question:    Why is the refractive index of atmosphere different at different altitudes? Ans: it is because it is a mixture of differnet particles and gas molecules which often has diff. refractive indices Question:    When a light ray passes obliquely through the atmosphere in an upward direction, how does its path generally change? Ans: When a light ray passes obliquely through the atmosphere in an upward direction, it 's path changes due to atmospheric refraction. This refraction happens because each atmospheric layer has a different optical density, due to which its path changes. Question:  ...