Focus and focal length - class-XII

Here $R _1$ = 20 cm, $R _2$ = -40 cm, f = 20 cm
Using lens maker's formula we get,
$\dfrac{1}{20} , = , (\mu , - , 1) \left ( \dfrac{1}{20} , + , \dfrac{1}{40} \right )$
$\dfrac{1}{20} , = , (\mu , - , 1) \dfrac{3}{40} , \Rightarrow , \mu , = , \dfrac{5}{3}$

The radius of curvature of a plane mirror is infinite so focal length will also be infinite.

since we know that radius of curvature is 2 times focal length i.e R=2f

Focal length of  spherical mirror is half the radius of curvature i.e f $=\cfrac { R }{ 2 } $ .

Focal length of  spherical mirror is half the radius of curvature i.e $f =\dfrac R2$

For spherical mirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. We put this as $R = 2f$. This implies that the principal focus of a spherical mirror lies midway between the pole and centre of curvature.

The radius of curvature of the mirror $=24 cm.$
Thus the focal length
$=24 cm/2$
$=12 cm.$

Lens maker's formula: 1/f = (n-1)(1/R1 - 1/R2). Given f=6, n=1.5, R1=x, R2=-2x (double-convex). 1/6 = (0.5)(1/x + 1/2x) = 0.5(3/2x) = 3/4x. So 4x = 18, x = 4.5 cm.

$A=20m$  $f=10m.$

Radius of curvature $=20 cm$

$\begin{array}{l} \dfrac { 1 }{ V } +\dfrac { 1 }{ \mu  } =\dfrac { 1 }{ f }  \ \dfrac { 1 }{ V } =\dfrac { 1 }{ { 10 } } ,V=10cm \end{array}$

First, find the image position from the lens: 1/v - 1/-15 = 1/10 => 1/v = 1/10 - 1/15 = 1/30. v = 30 cm. For the mirror to make the image coincide, the light must strike the mirror normally, meaning the image must be at the center of curvature. Mirror R = 12, so C = 12. Distance = 30 - 12 = 18 cm.

Magnification m = h_i/h_o = -v/u. m = -5/10 = -0.5. -0.5 = -(-30)/u => u = -60. Mirror formula: 1/f = 1/v + 1/u = 1/-30 + 1/-60 = -3/60 = -1/20. f = -20 cm.

A convex mirror always forms a virtual, erect, and diminished image for any real object position. Therefore, the magnification is always less than 1. Statement A is correct. Statement D is incorrect because a convex mirror cannot form an inverted image for a real object.

The focal length of a spherical mirror depends solely on its radius of curvature and is independent of the medium in which it is placed. Therefore, immersing a spherical mirror in a liquid does not change its focal length, unlike lenses whose focal length depends on refractive indices.

For a mirror, the magnification m = -v/u = f/(f-u). Given the sun is very far away (u is large), the image is formed at the focus f. The size of the image is h_i = h_o * (f/u). Plugging in values: (1.4 * 10^9) * (2 / 10^11) = 2.8 * 10^-2 m = 2.8 cm.

For a concave mirror, magnification m = f / (f - u). Rearranging for u: m(f - u) = f, mf - mu = f, mu = mf - f, u = f(m - 1) / m. However, for a real image, m is negative. Using m = -|m|, the distance u = f(1 + |m|) / |m|.

Using the mirror formula 1/v + 1/u = 1/f. For converging rays, u = +30 cm (virtual object). The reflected rays form a diverging beam with image at v = -120 cm. 1/(-120) + 1/30 = 1/f. 1/f = (-1 + 4) / 120 = 3/120 = 1/40. So f = 40 cm.

Rear view mirrors in vehicles are convex mirrors. They are designed to provide a wider field of view by forming diminished images, so the magnification is always less than one.

Mirror operates on the principle of Laws of reflection. 

Plane mirrors are not curved.

Convex lens gives the real and double-sized image when the object is placed exactly between the focus and radius of curvature.
We have, $\displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$m = \displaystyle \frac{v}{u} = 2$ or $v = 2u$

The relation between focal length and Radius of curvature is,

Image of an object placed at center of curvature is inverted, real and of the same size and is formed at the center of curvature. Hence image will be real and at the center of curvature (20 cm from mirror).

The standard lens formula is 1/v - 1/u = 1/f. Solving for f gives f = (uv) / (u - v).

For the plane mirror, the image is formed 30 cm behind it. For the convex mirror, the object is at 50 cm. No parallax means the images coincide. This setup requires calculating the focal length based on the image position.