An Object Is Placed 14 Cm In Front Of A Concave Mirror With A 57 Cm Focal Length. Where Is The

An object is placed 4 cm from a concave mirror of focal length 6 cm. What is the position, magnification, and nature of the image formed?

distance of object from mirror=p=4cmfocal length=6cmUsing mirror formula1/f=1/p+1/q1/6=1/4+1/q1/6–1/4=1/q1/q=(2–3)/121/q=-1/12q=-12PositiOn of image is 12cm from mirrorNature of Image is virtual, erect and magnified(M=3).

An object is placed in front of a concave mirror of focal length 10 cm. Where should a screen be placed to obtain an image double the size of object?

The mirror equation is [math]\frac{1}{v}+\frac{1}{u}=\frac{1}{f},[/math] where [math]v, u[/math] and [math]f[/math] are the distances of the image and object from the mirror and the focal length respectively.Magnification, [math]M = \frac{h_i}{h_o} = -\frac{v}{u},[/math] where [math]h_i[/math] and [math]h_o[/math] are the heights of the image and object respectively.As per the new Cartesian convention, distances from the mirror toward the side of the object are negative and distance toward the other side are positive. Distances above the principal axis are positive and distances below the principal axis are negative.It is given that the image, as obtained on the screen, is double the size of the object.[math]\Rightarrow \qquad[/math] The image is real and hence is inverted.[math]\Rightarrow \qquad h_i=-2h_o \qquad \Rightarrow \qquad \frac{h_i}{h_o}=-2.[/math][math]\Rightarrow \qquad -\frac{v}{u}=-2 \qquad \Rightarrow \qquad v=2u.[/math]It is also given that the focal length is [math]10[/math] cm.[math]\Rightarrow \qquad f=-10 \qquad[/math] (In a concave mirror, [math]f[/math] is negative).[math]\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \qquad \Rightarrow \qquad \frac{1}{2u}+\frac{1}{u}=-\frac{1}{10}.[/math][math]\Rightarrow \qquad \frac{3}{2u}=-\frac{1}{10} \qquad \Rightarrow \qquad u=-15 \qquad \Rightarrow \qquad v=-30.[/math][math]\Rightarrow \qquad[/math] The screen should be placed at a distance of [math]30[/math] cm from the mirror.

An object is placed 30 cm in front of a concave mirror of focal length 20 cm. What is the distance between the object and the image?

Given,u=30 cmf=20 cmv=?Here,1/u + 1/v = 1/f1/30 + 1/v = 1/201/v = 1/20 - 1/301/v = 1/60 (use calculator )v= 60 cm( u, v, f are always in ‘+’ in concave.But ‘-’ in convex)Now,(Since image is formed on the same side of the object , the image is real)v-u = 60 - 30 = 30 cmSo the distance between object and image is 30 cm.

An object is placed at the distance of 10 cm in front of concave mirror and its focal length is 20 cm, can you find the position of image formed?

1/u + 1/v = 1/fHere 1/10 + 1/20 = 3/20 = 1/ff = 20/31/v = 1/f – 1/u = (u – f)/ufv = uf/(u-f)dv/du = [(u – f)*f – uf]/(u-f)2-quotient rule = -f2/(u-f)2[dv/du]f=20/3,u=10= 4Δv = [dv/du]f=20/3,u=10* Δu =4*0.1 = 0.4 therefore the image will shift by 0.4 cm //

The focal length of a concave mirror is 50 cm. Where can an object be placed so that its image is magnified two times and inverted?

f=50cmv=2uu=uNow use 1/f=1/v+1/uU will get 75cm which is the ans

An object of size 7 cm is placed at 27 cm in front of a concave mirror of focal length is 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained?

GivenFocal length of concave mirror (f)= -18cmConventionally concave mirror and concave lens have negative focal lengths and convex mirror and convex lens have positive focal lengths. The co-ordinate system is taken with respect to the Pole as origin. All distances are measured from the pole and distance measured in the direction of light rays are positive and opposite to the direction are considered as negative. Distances above the principal axis are positive and vice-versa. Following these conventionsObject distance (u)= -27cmObject height= 7cmLet image distance be v cmWe know by mirror formula that1/v + 1/u = 1/f1/v + 1/(-27) = 1/(-18)1/v = 1/(-18) + 1/271/v= (-3+2)/541/v= -1/54v= -54 cmThus, placing a screen at a distance of 54 cm from the concave mirror on the same side as object will give us a sharp and focused image.The magnification will bem= -v/um= -(-54)/(-27)m= -2|m|>1 => enlarged image and the negative sign implies that the image is invertedm can also be written asm= Height of image/ Height of object-2= Height of image/7Height of image= -14cmThus the image has a height of 14 cm and is inverted and present at a distance of 54 cm from the mirror.