Image Formation by Converging Lenses - Class XII

A virtual image is formed when the refracting or reflecting rays do not actually intersect but appear to intersect when they are produced backwards.

The image which can be obtained on a screen is called a real image.

A virtual image is always erect.

Real images can be obtained on a screen. Virtual images can not be obtained on a screen.

When the refracting or reflecting rays actually intersect, a real image is formed.

A real image is always inverted.

The convex lens forms an image at its focal point (30 cm). The concave lens is placed 26 cm from the convex lens, so the image acts as a virtual object for the concave lens at a distance of 4 cm (30 - 26 = 4 cm). Using the lens formula 1/f = 1/v - 1/u, where f = -20 and u = +4, we get 1/v = 1/-20 + 1/4 = 4/20, so v = 5 cm. Magnification m = v/u = 5/4 = 1.25. Final height = 1.25 * 2 cm = 2.50 cm.

Using the lens formula 1/f = 1/v - 1/u with f = 40 cm, the near end at u1 = -60 cm has an image distance v1 = -120 cm. The far end of the meter stick is at u2 = -160 cm, leading to an image distance v2 = -53.33 cm. The length of the image is the difference between these image positions, which evaluates to 200/3 cm.

In the displacement method, the object size O is related to the two image sizes I1 and I2 by the geometric mean formula O = sqrt(I1 * I2). This result is derived from the magnification formulas m1 = I1/O = v/u and m2 = I2/O = u/v.

Similar to the previous question, the size of the object in the displacement method is the geometric mean of the two image sizes, O = sqrt(I1 * I2).

The given problem is an example of displacement method, which is generally used to measure the focal length of the lens. In this method, the two image sizes and the object size are related as:

When a lens is rotated about its optical center, the image of a point object on the principal axis moves in a circular arc centered at the optical center. The new image position P will not lie on either the original or the new principal axis.

Using the displacement method formulas: m1 = 25, m2 = 1/25 (since the lens is moved). The distance between positions is d = 30 cm. The formula for focal length is f = (D^2 - d^2) / 4D. Alternatively, using magnification m = (D-d)/2f, one can solve for f.

$h^{2} _{object}=h _{image1} \times h _{image2}$

For a lens with object-image distance 4f, the magnification m = -1 at the center. Near this point, the displacement of the image is greater than the displacement of the object (m > 1 or m < -1), but the question asks about relative movements. Specifically, for a real object and real image, the longitudinal magnification is m^2. Since m^2 > 1 for positions away from 2f, the image moves more than the object.

The lens maker formula states that (1/f) = (n - 1)(1/R1 - 1/R2). For the violet rays, 1/25 = (1.55 - 1) * K, where K represents the curvature term (1/R1 - 1/R2). Solving gives K = 1 / (0.55 * 25). For red rays, 1/f_red = (1.50 - 1) * K = 0.50 * [1 / (0.55 * 25)], which yields f_red = 27.5 cm.

The image size h_i = f * tan(theta), where theta is the angular diameter of the sun. tan(theta) = diameter / distance = 1.4e9 / 1.5e11 = 1.4/150. h_i = 20 cm * (1.4/150) = 200 mm * 0.00933 = 1.86 mm. The radius is half of this, 0.93 mm.

Using 1/v - 1/u = 1/f with u = -8 cm and f = 12 cm, 1/v = 1/12 - 1/8 = (2-3)/24 = -1/24. So v = -24 cm. Magnification m = v/u = -24 / -8 = 3. Image size = m * object size = 3 * 2.0 cm = 6 cm.

For a thin lens, an image of the same size as the object (magnification m = -1) is formed when the object is at 2f from the lens.

The object is at 2f (40 cm), so the image is at 2f (40 cm) on the other side. The lens diameter is 10 cm (radius 5 cm). The rays from the object pass through the lens and converge at the image point. The cone of light has a radius that scales with distance. At 60 cm from the lens (20 cm past the image), the cone radius is determined by similar triangles.

$height^{2} _{object}=height _{image1}\times height _{image2}$

$height^{2} _{object}=height _{image1} \times height _{image2}$

Initial positions: candle 12, lens 50, screen 88. Object distance u = 50 - 12 = 38 cm. Image distance v = 88 - 50 = 38 cm. Since u = v, 2f = 38, so f = 19 cm. If the candle is moved to 31 cm, the new object distance u = 50 - 31 = 19 cm. Since u = f, the image is formed at infinity.

For a concave lens, object distance u = -30 cm and focal length f = -15 cm. Using the lens formula 1/f = 1/v - 1/u, we find v = -10 cm. The magnification is m = v/u = -10 / -30 = 1/3. Since magnification is also height of image / height of object, and image height is 3 cm, the object height must be 3 cm divided by (1/3), which equals 9 cm.

Magnification of area = 900 times

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$u= -u$ ; $f= f$

When an object is placed between $F$ and $2F$ then image will formed between $F$ and $2F$ on opposite side of lens  and Image formed is real, Inverted and 

Answer is A.

The expression for focal length by displacement method is given as follows.
$f=\frac { { D }^{ 2 }-{ x }^{ 2 } }{ 4D } $
where,
D - the distance between the object and screen
x - the distance between the two positions of the lens.
Here, D = 100 cm and x = 40 cm.
So, $f=\frac { { D }^{ 2 }-{ x }^{ 2 } }{ 4D } =\frac { { 100 }^{ 2 }-{ 40 }^{ 2 } }{ 4\times 100 } =21\quad cm$.
Hence, the focal length of the lens is 21 cm.

If the image sizes be $ {I} _{1} \  and \  {I} _{2} $,

Here $u _1 + u _2  =-24$.......(1).

Given, magnification $=-\dfrac{v}{u}=3$ and $v=80\ cm$
So, object distance, $u=-\dfrac{v}{3}=-\dfrac{80}{3}=-26.67\ cm$

From lens formula, $\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Given $\displaystyle \frac{v}{u} = 2.5$
$v= 2.5 u$
again $ v-u = 30$
$v=30+u$
$2.5 u = 30 +u$
$1.5 u=30$
$u=20$
Now, $\displaystyle \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{u+v}{uv}$
$f = \displaystyle \frac{uv}{u+v}= \frac{2.5 u^2}{3.5 u}$
   $\displaystyle =\frac{5}{7}u = \frac{5 \times 20}{7} = \frac{100}{7}=14.3cm$