Question

A convex lens forms an image 16.0 cm long of an object 4.0 cm long kept at a distance of 6 cm from the lens. The object and the image are on the same side of the lens.
(a) What is nature of the image?
(b) Find (i) the position of the image, and (ii) the focal length of the lens.​

Answer 1

Answer:

Need to find: The three angles of ∆?

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❍ Let's say the smaller angle be x. Then, the largest angle and another angle would be (4x + 5) and 2x.

$\begin{gathered}\underline{\bigstar\:{\pmb{\textsf{Angle Sum Property of}} \: \sf{\Delta} \;:}}\\\end{gathered}$

The ASP (Angle Sum Property) of the triangle States that the Sum of all angles of the triangle is 180°.

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$\begin{gathered}:\implies\sf x + 4x + 5 + 2x = 180^\circ\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 7x + 5 = 180^\circ\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 7x = 180^\circ - 5\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 7x = 175\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf x = \cancel\dfrac{175^\circ}{7}\\\\\\\end{gathered}$

$\begin{gathered}:\implies\underline{\boxed{\pmb{\frak{\purple{x = 25^\circ}}}}}\;\bigstar\\\end{gathered}$

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$\underline{\bf{\dag} \:\mathfrak{Angles\;of\;the\:\Delta\:are\; :}}$

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x = 25°

(4x + 5) = (4[25] + 5) = 105°

2x = 2(25) = 50°

⠀

$\therefore{\underline{\sf{Hence, \: the \: angles \: of \: \Delta \: are \: \pmb{ 25^\circ, \; 105^\circ,\; 50^\circ} \: respectively}}}$

$\huge\mathtt\red{\textsf{MissEleGant}}$

Answer 2

Answer :-

Given:-

Image height (I) = 16.0 cm (positive)

Object height(0) = 4.0 cm (positive)

Object distance (u) = 6 cm (negative)

(i) From relation:

⟹ m = I/O = v/u

Where:-

m = magnification

I = image height

O = object height

u = object distance

v = image distance

Since:-

⟹ m = I/0 = v/u = 160/40 = v/-6

or v = -24 cm

Thus, the image is at a distance 24 cm in front of the lens.

(ii) From relation:

⟹ 1/v - 1/u = 1/f

Where:-

v = image distance

u = object distance

f = focal length

Since:-

⟹ 1/f = 1/-24 - 1/(-6)

⟹ 1/f = -1/24 + 1/6 = 1/8

or f = 8 cm

Thus, the focal length of the lens is 8 cm.

Nature of the image:-

Since, the image is magnified and on the same side of the lens as the object, so the image is virtual.